RD Sharma Solutions for Class 10 Math Chapter 6 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among class 10 students for Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma Book of class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma Solutions. All RD Sharma Solutions for class 10 Math are prepared by experts and are 100% accurate.
Page No 6.42:
Question 1:
Prove the following trigonometric identities.
(1 − cos2 A) cosec2 A = 1
Answer:
We know that,
So,
Page No 6.42:
Question 2:
Prove the following trigonometric identities.
(1 + cot2 A) sin2 A = 1
Answer:
We know that,
So,
Page No 6.42:
Question 3:
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
Answer:
We know that, .
So,
Page No 6.42:
Question 4:
Prove the following trigonometric identities.
Answer:
We know that,
So,
Page No 6.42:
Question 5:
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Answer:
We know that,
So,
Page No 6.42:
Question 6:
Prove the following trigonometric identities.
Answer:
We know that,
So,
Page No 6.42:
Question 7:
Prove the following trigonometric identities.
Answer:
We know that,
Multiplying both numerator and the denominator by, we have
Page No 6.42:
Question 8:
Prove the following trigonometric identities.
Answer:
We know that,
Multiplying the both numerator and the denominator by, we have
Page No 6.42:
Question 9:
Prove the following trigonometric identities.
Answer:
We know that,
So,
Page No 6.42:
Question 10:
Prove the following trigonometric identities.
Answer:
We know that,
So,
Page No 6.42:
Question 11:
Prove the following trigonometric identities.
Answer:
We know that,
Multiplying numerator and denominator under the square root by , we have
Page No 6.42:
Question 12:
Prove the following trigonometric identities.
Answer:
We have to prove .
We know that,
Multiplying both numerator and denominator by , we have
Page No 6.42:
Question 13:
Prove the following trigonometric identities.
Answer:
We have to prove .
We know that,
Multiplying both numerator and denominator by , we have
Page No 6.42:
Question 14:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that, .
Multiplying both numerator and denominator by , we have
Page No 6.43:
Question 15:
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2θ + cos2θ
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 16:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 17:
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Answer:
We have to prove
We know that,
Page No 6.43:
Question 18:
Prove the following trigonometric identities.
secA (1 − sinA) (secA + tanA) = 1
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 19:
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 20:
Prove the following trigonometric identities.
tan2θ − sin2θ = tan2θ sin2θ
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 21:
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 22:
Prove the following trigonometric identities.
sin2A cot2A + cos2A tan2 A = 1
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 23:
Prove the following trigonometric identities.
(i)
(ii)
Answer:
(i) We have to prove
We know that,
So,
(ii) We have to prove
We know that,
So,
Page No 6.43:
Question 24:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 25:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that,
So,
Page No 6.43:
Question 26:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that,
Multiplying the denominator and numerator of the second term by , we have
Page No 6.43:
Question 27:
Prove the following trigonometric identities.
Answer:
We have to prove that .
We know that,
So,
Page No 6.43:
Question 28:
Prove the following trigonometric identities.
Answer:
We have to prove
Consider the expression
Again, we have
Page No 6.43:
Question 29:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that,
Multiplying the numerator and denominator by , we have
Page No 6.43:
Question 30:
Prove the following trigonometric identities.
Answer:
We need to prove
Now, using in the L.H.S, we get
Further using the identity, we get
Hence
Page No 6.43:
Question 31:
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
Answer:
We need to prove
Solving the L.H.S, we get
Further using the identity, we get
Hence proved.
Page No 6.43:
Question 32:
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Answer:
We need to prove
Solving the L.H.S, we get
()
Further using the identity, we get
Hence proved.
Page No 6.43:
Question 33:
Prove the following trigonometric identities.
Answer:
We need to prove
Solving the L.H.S, we get
Using , we get
Hence proved.
Page No 6.43:
Question 34:
Prove the following trigonometric identities.
Answer:
We need to prove
Using the property , we get
LHS =
Further using the identity, , we get
= RHS
Hence proved.
Page No 6.43:
Question 35:
Prove the following trigonometric identities.
Answer:
We need to prove
Here, we will first solve the LHS.
Now, using , we get
Further, multiplying both numerator and denominator by , we get
Now, using the property, we get
So,
= RHS
Hence proved.
Page No 6.43:
Question 36:
Prove the following trigonometric identities.
Answer:
We need to prove
Now, multiplying the numerator and denominator of LHS by , we get
Further using the identity, , we get
Hence proved.
Page No 6.43:
Question 37:
Prove the following trigonometric identities.
(i)
(ii)
Answer:
(i) We need to prove
Here, rationalising the L.H.S, we get
Further using the property, , we get
So,
Hence proved.
(ii) We need to prove
Here, rationaliaing the L.H.S, we get
Further using the property, , we get
So,
Hence proved.
Page No 6.43:
Question 38:
Prove the following trigonometric identities.
Answer:
We need to prove
Here, rationaliaing the L.H.S, we get
Further using the property, , we get
So,
Hence proved.
Page No 6.43:
Question 39:
Prove the following trigonometric identities.
Answer:
We need to prove
Here, we will first solve the L.H.S.
Now, using , we get
Further using the property , we get
So,
Hence proved.
Page No 6.43:
Question 40:
Prove the following trigonometric identities.
Answer:
We need to prove
Now, rationalising the L.H.S, we get
Using and, we get
Hence proved.
Page No 6.43:
Question 41:
Prove the following trigonometric identities.
Answer:
We need to prove
Solving the L.H.S, we get
Further using the property , we get
So,
Hence proved.
Page No 6.43:
Question 42:
Prove the following trigonometric identities.
Answer:
We need to prove
Solving the L.H.S, we get
= RHS
Hence proved.
Page No 6.43:
Question 43:
Prove the following trigonometric identities.
Answer:
We need to prove
Using the identity, we get
Further, using the property , we get
So,
Hence proved.
Page No 6.44:
Question 44:
Prove the following trigonometric identities.
Answer:
We need to prove .
Using the property , we get
Now, using , we get
Further, using the property, , we get
Hence proved.
Page No 6.44:
Question 45:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove .
Here, we will first solve the LHS.
Now, using , we get
On further solving by taking the reciprocal of the denominator, we get,
Hence proved.
Page No 6.44:
Question 46:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Here, we will first solve the LHS.
Now, using, we get
On further solving by taking the reciprocal of the denominator, we get,
Now, taking common from both the numerator and the denominator, we get
Hence proved.
Page No 6.44:
Question 47:
Prove the following trigonometric identities.
(i)
(ii)
(iii)
(iv)
Answer:
(i) We have to prove the following identity-
Consider the LHS.
= RHS
Hence proved.
(ii) We have to prove the following identity-
Consider the LHS.
(Divide numerator and denominator by)
RHS
Hence proved.
(iii) We have to prove the following identity-
Consider the LHS.
= RHS
Hence proved.
(iv)
Consider the LHS.
= RHS
Hence proved.
Page No 6.44:
Question 48:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove .
Here, we will first solve the L.H.S.
Now, using, we get
On further solving, we get
Similarly we solve the R.H.S.
Now, using, we get
On further solving, we get
So, L.H.S = R.H.S
Hence proved.
Page No 6.44:
Question 49:
Prove the following trigonometric identities.
tan2A + cot2A = sec2A cosec2A − 2
Answer:
In the given question, we need to prove
Now, using and in L.H.S, we get
Further, using the identity, we get
Since L.H.S = R.H.S
Hence proved.
Page No 6.44:
Question 50:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Now, using and in the L.H.S, we get
Solving further, we get
Hence proved.
Page No 6.44:
Question 51:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Using and, we get
Further, using the property , we get
Hence proved.
Page No 6.44:
Question 52:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Using the identity, we get
Further, using the property , we get
Hence proved.
Page No 6.44:
Question 53:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove .
Using the property , we get
So,
Solving further, we get
Hence proved.
Page No 6.44:
Question 54:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Using the property and, we get
Taking the reciprocal of the denominator, we get
Further, using the identity, we get
Hence proved.
Page No 6.44:
Question 55:
Prove the following trigonometric identities.
Answer:
In the given question, we are given
We need to prove
Here L.H.S is
Now, solving the L.H.S, we get
Further using the property, we get
So,
Now, solving the R.H.S, we get
So,
Further using the property, we get,
So,
Hence proved.
Page No 6.44:
Question 56:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Now, using the identity in L.H.S, we get
Further using, we get
Also, from the identity , we get
Hence proved.
Page No 6.44:
Question 57:
Prove the following trigonometric identities.
Answer:
In the given question, we need to prove
Now, using and in L.H.S, we get
Further using the identity, we get
Further using the identity , we get
Now, from the identity, we get
So,
Hence proved.
Page No 6.44:
Question 58:
Prove the following trigonometric identities.
(i)
(ii)
Answer:
(i) In the given question, we need to prove
Taking common from the numerator and the denominator of the L.H.S, we get
Now, using the property , we get
Using , we get
Taking common from the numerator, we get
Using and , we get
Now, using the property , we get
Hence proved.
(ii)
Consider the LHS.
= RHS
Hence proved.
Page No 6.44:
Question 59:
Prove the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Answer:
We have to prove
We know that,
So, we have
So, we have
Hence proved.
Page No 6.44:
Question 60:
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
Answer:
We have to prove
We know that, .
So,
Hence proved.
Page No 6.44:
Question 61:
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Answer:
We have to prove
We know that,
Consider the LHS.
Now, consider the RHS.
∴ LHS = RHS
Hence proved.
Page No 6.44:
Question 62:
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Answer:
We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 63:
Prove the following trigonometric identities.
(i)
(ii)
Answer:
(i) We have to prove
So,
Hence proved.
(ii) We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 64:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 65:
Prove the following trigonometric identities.
Answer:
We have to prove
We know that, .
So,
Hence proved.
Page No 6.45:
Question 66:
Prove the following trigonometric identities.
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Answer:
We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 67:
Prove the following trigonometric identities.
Answer:
We have to prove .
We know that,
So,
Multiplying both the numerator and denominator by , we have
Hence proved.
Page No 6.45:
Question 68:
Prove the following trigonometric identities.
Answer:
We have prove that
We know that,
So,
Now,
Hence proved.
Page No 6.45:
Question 69:
Prove the following trigonometric identities.
sin2A cos2B − cos2A sin2B = sin2A − sin2B
Answer:
We know that,
So have,
Hence proved.
Page No 6.45:
Question 70:
Prove the following trigonometric identities.
(i)
(ii)
Answer:
(i) We have to prove
Now,
Hence proved.
(ii) We have to prove
Now,
Hence proved.
Page No 6.45:
Question 71:
Prove the following trigonometric identities.
Answer:
We have to prove
Now,
Hence proved.
Page No 6.45:
Question 72:
Prove the following trigonometric identities.
cot2 A cosec2 B − cot2 B cosec2 A = cot2 A − cot2 B
Answer:
We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 73:
Prove the following trigonometric identities.
tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2B
Answer:
We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 74:
Prove the following trigonometric identities.
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
Answer:
Given that,
We have to prove
We know that,
So,
Hence proved.
Page No 6.45:
Question 75:
Prove the following trigonometric identities.
If
Answer:
Given that,
We have to prove
We know that,
Squaring and then adding the above two equations, we have
Page No 6.45:
Question 76:
Prove the following trigonometric identities.
If cosec θ − sin θ = a3, sec θ − cos θ = b3, prove that a2b2 (a2 + b2) = 1
Answer:
Given that,
We have to prove
We know that
Now from the first equation, we have
Again from the second equation, we have
Therefore, we have
Hence proved.
Page No 6.45:
Question 77:
Prove the following trigonometric identities.
If a cos3 θ + 3 a cos θ sin2 θ = m, a sin3 θ + 3 a cos2 θ sin θ = n, prove that (m + n)2/3 + (m − n)2/3 = 2a2/3
Answer:
Given that,
We have to prove
Adding both the equations, we get
Also.
Therefore, we have
Hence proved.
Page No 6.45:
Question 78:
Prove the following trigonometric identities.
If x = a cos3 θ, y = b sin3 θ, prove that
Answer:
Given:
We have to prove
We know that
So, we have
Hence proved.
Page No 6.45:
Question 79:
Prove the following trigonometric identities.
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ − 3 cos θ = ± 3.
Answer:
Given:
We have to prove that 5sin θ – 3cos θ = ±3.
We know that,
Squaring the given equation, we have
⇒ (5sin θ – 3cos θ)2 = 9
⇒ 5sin θ – 3cos θ = ±3
Hence proved.
Page No 6.45:
Question 80:
Prove the following trigonometric identities.
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that a2 + b2 = m2 + n2
Answer:
Given:
We have to prove
We know that,
Now, squaring and adding the two equations, we get
Hence proved.
Page No 6.45:
Question 81:
Prove the following trigonometric identities.
If cosec θ + cot θ = m and cosec θ − cot θ = n, prove that mn = 1
Answer:
Given:
We have to prove
We know that,
Multiplying the two equations, we have
Hence proved.
Page No 6.45:
Question 82:
Prove the following trigonometric identities.
if cos A + cos2A = 1, prove that sin2 A + sin4 A = 1
Answer:
Given:
We have to prove
Now,
Therefore, we have
Hence proved.
Page No 6.45:
Question 83:
Prove the following trigonometric identities.
(i)
(ii)
(iii)
(iv)
(v)
Answer:
(i) We have,
(ii) We have,
(iii) We have,
(iv) We have,
Multiplying both the numerator and the denominator by, we have
(v) We have,
Multiplying both the numerator and the denominator by , we have
Page No 6.46:
Question 84:
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Answer:
Given:
We have to prove
From the given equation, we have
Therefore, we have
Hence proved.
Page No 6.46:
Question 85:
Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
Show that one of the values of each member of this equality is sin α sin β sin γ
Answer:
Given:
Let us assume that
We know that,
Then, we have
Therefore, we have
Taking the expression with the positive sign, we have
Page No 6.46:
Question 86:
If sin θ + cos θ = x, prove that .
Answer:
Given:
Squaring the given equation, we have
Squaring the last equation, we have
Therefore, we have
Hence proved.
Page No 6.46:
Question 87:
If x = a sec θ cos Ï, y = b sec θ sin Ï and z = c tan θ, show that .
Answer:
Given:
We have to prove that .
Squaring the above equations and then subtracting the third from the sum of the first two, we have
Hence proved.
Page No 6.53:
Question 1:
If , find all other trigonometric ratios of angle θ.
Answer:
Given:
Now, we have to find all the other trigonometric ratios.
We have the following right angle triangle.
From the above figure,
Therefore,
Page No 6.53:
Question 2:
If , find all other trigonometric ratios of angle θ.
Answer:
Given:
We have to find all the trigonometric ratios.
We have the following right angle triangle.
From the above figure,
Page No 6.53:
Question 3:
If , find the value of .
Answer:
Given:
We have to find the value of the expression
We know that,
Therefore, the given expression can be written as
Hence, the value of the given expression is .
Page No 6.53:
Question 4:
If , find the value of .
Answer:
Given:
We have to find the value of the expression .
From the above figure, we have
Therefore,
Hence, the value of the given expression is .
Page No 6.53:
Question 5:
If , find the value of .
Answer:
Given:
We have to find the value of the expression .
From the above figure, we have
Therefore,
Hence, the value of the given expression is 25.
Page No 6.53:
Question 6:
If , find the value of .
Answer:
Given:
We have to find the value of the expression
We know that,
Using the identity , we have
Hence, the value of the given expression is .
Page No 6.53:
Question 7:
If , find the value of .
Answer:
Given:
We have to find the value of the expression
We know that,
Therefore,
Hence, the value of the given expression is 2.
Page No 6.53:
Question 8:
If , find the value of .
Answer:
Given:
We have to find the value of the expression .
We know that,
Therefore,
Hence, the value of the given expression is .
Page No 6.53:
Question 9:
If 3cosθ = 1, find the value of
Answer:
Given:
We have to find the value of the expression .
We have,
Therefore,
Hence, the value of the expression is 10.
Page No 6.53:
Question 10:
If , find the value of sin2θ − cos2θ.
Answer:
Given:
We have to find the value of .
Therefore,
Hence, the value of the expression is .
Page No 6.53:
Question 11:
If , find the value of .
Answer:
Given:
We have to find the value of the expression .
Now,
Therefore,
Hence, the value of the expression is 3.
Page No 6.53:
Question 12:
If sin θ + cos θ = , find cot θ.
Answer:
Given:
We have to find the value of .
Now,
Hence,
Page No 6.54:
Question 1:
Define an identity.
Answer:
An identity is an equation which is true for all values of the variable (s).
For example,
Any number of variables may involve in an identity.
An example of an identity containing two variables is
The above are all about algebraic identities. Now, we define the trigonometric identities.
An equation involving trigonometric ratios of an angle (say) is said to be a trigonometric identity if it is satisfied for all valued of for which the trigonometric ratios are defined.
For examples,
Page No 6.54:
Question 2:
What is the value of (1 − cos2 θ) cosec2 θ?
Answer:
We have,
Page No 6.54:
Question 3:
What is the value of (1 + cot2 θ) sin2 θ?
Answer:
We have,
Page No 6.54:
Question 4:
What is the value of ?
Answer:
We have,
Page No 6.54:
Question 5:
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Answer:
Given:
We know that,
Therefore,
Hence,
Page No 6.54:
Question 6:
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
Answer:
Given:
We know that,
Therefore,
Hence,
Page No 6.54:
Question 7:
Write the value of cosec2 (90° − θ) − tan2 θ.
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 8:
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 9:
Write the value of .
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 10:
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
Answer:
Given:
x = a sinθ and y = b cosθ
So,
We know that,
Therefore,
Page No 6.54:
Question 11:
If , what is the value of cotθ + cosecθ?
Answer:
Given:
We know that,
We have,
Hence, the value of cotθ + cosecθ is 2.Page No 6.54:
Question 12:
What is the value of 9cot2 θ − 9cosec2 θ?
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 13:
What is the value of .
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 14:
What is the value of .
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 15:
What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?
Answer:
We have,
We know that,
Therefore,
Page No 6.54:
Question 16:
If , find the value of tan A + cot A.
Answer:
Given:
We know that,
Therefore,
Page No 6.54:
Question 17:
If , then find the value of 2cot2 θ + 2.
Answer:
Given:
We know that,
Therefore,
Page No 6.54:
Question 18:
If , then find the value of 9tan2 θ + 9.
Answer:
Given:
We know that,
Therefore,
Page No 6.54:
Question 19:
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
Answer:
Given:
Hence, the value of k is 1.
Page No 6.54:
Question 20:
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Answer:
Given:
Thus, the value of λ is 1.Page No 6.54:
Question 21:
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Answer:
Given:
Hence, the value of λ is 1.
Page No 6.54:
Question 22:
If 5x = sec θ and , find the value of .
Answer:
Given:
We know that,
Page No 6.55:
Question 23:
If cosec θ = 2x and , find the value of .
Answer:
Given:
We know that,
Page No 6.55:
Question 1:
If sec θ + tan θ = x, then sec θ =
(a)
(b)
(c)
(d)
Answer:
Given:
We know that,
Now,
Adding the two equations, we get
Therefore, the correct choice is (b).
Page No 6.55:
Question 2:
If , then
(a)
(b)
(c)
(d)
Answer:
Given:
We know that,
Now,
Subtracting the second equation from the first equation, we get
Therefore, the correct choice is (d).
Page No 6.55:
Question 3:
is equal to
(a) sec θ + tan θ
(b) sec θ − tan θ
(c) sec2 θ + tan2 θ
(d) sec2 θ − tan2 θ
Answer:
The given expression is .
Multiplying both the numerator and denominator under the root by , we have
Therefore, the correct option is (a).
Page No 6.55:
Question 4:
The value of is
(a) cot θ − cosec θ
(b) cosec θ + cot θ
(c) cosec2 θ + cot2 θ
(d) (cot θ + cosec θ)2
Answer:
The given expression is .
Multiplying both the numerator and denominator under the root by, we have
Therefore, the correct choice is (b).
Page No 6.55:
Question 5:
sec4 A − sec2 A is equal to
(a) tan2 A − tan4 A
(b) tan4 A − tan2 A
(c) tan4 A + tan2 A
(d) tan2 A + tan4 A
Answer:
The given expression is .
Taking common from both the terms, we have
Disclaimer: The options given in (c) and (d) are same by the commutative property of addition.
Therefore, the correct options are (c) or (d).
Page No 6.55:
Question 6:
cos4 A − sin4 A is equal to
(a) 2 cos2 A + 1
(b) 2 cos2 A − 1
(c) 2 sin2 A − 1
(d) 2 sin2 A + 1
Answer:
The given expression is .
Factorising the given expression, we have
Therefore, the correct option is (b).
Page No 6.55:
Question 7:
is equal to
(a)
(b)
(c)
(d)
Answer:
The given expression is .
Multiplying both the numerator and denominator under the root by , we have
Therefore, the correct option is (c).
Page No 6.56:
Question 8:
is equal to
(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ − cos θ
Answer:
The given expression is .
Simplifying the given expression, we have
Therefore, the correct option is (c).
Page No 6.56:
Question 9:
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
(a) 1
(b) 2
(c) 4
(d) 0
Answer:
The given expression is
Simplifying the given expression, we have
Therefore, the correct option is (b).
Page No 6.56:
Question 10:
is equal to
(a) 2 tan θ
(b) 2 sec θ
(c) 2 cosec θ
(d) 2 tan θ sec θ
Answer:
The given expression is .
Simplifying the given expression, we have
Therefore, the correct option is (c).
Page No 6.56:
Question 11:
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
(a) 0
(b) 1
(c) −1
(d) None of these
Answer:
The given expression is
Simplifying the given expression, we have
Therefore, the correct option is (b).
Page No 6.56:
Question 12:
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
(a) a2b2
(b) ab
(c) a4b4
(d) a2 + b2
Answer:
Given:
So,
We know that,
Therefore,
Hence, the correct option is (a).
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Question 13:
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
(a) ab
(b) a2 − b2
(c) a2 + b2
(d) a2b2
Answer:
Given:
So,
We know that,
Therefore,
Hence, the correct option is (d).
Page No 6.56:
Question 14:
is equal to
(a) 0
(b) 1
(c) −1
(d) 2
Answer:
The given expression is .
Simplifying the given expression, we have
Therefore, the correct option is (b).
Page No 6.56:
Question 15:
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
(a) 0
(b) 1
(c) −1
(d) None of these
Answer:
The given expression is .
Simplifying the given expression, we have
Therefore, the correct option is (c).
Page No 6.56:
Question 16:
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
(a) 7
(b) 12
(c) 25
(d) None of these
Answer:
Given:
Squaring and then adding the above two equations, we have
Hence, the correct option is (c).
Page No 6.56:
Question 17:
If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2 − q2 =
(a) a2− b2
(b) b2 − a2
(c) a2 + b2
(d) b − a
Answer:
Given:
Squaring both the equations and then subtracting the second from the first, we have
Hence, the correct option is (b).
Page No 6.56:
Question 18:
The value of sin2 29° + sin2 61° is
(a) 1
(b) 0
(c) 2 sin2 29°
(d) 2 cos2 61°
Answer:
The given expression is .
Hence, the correct option is (a).
Page No 6.56:
Question 19:
If x = r sin θ cos Ï, y = r sin θ sin Ï and z = r cos θ, then
(a)
(b)
(c)
(d)
Answer:
Given:
Squaring and adding these equations, we get
Hence, the correct option is (a).
Page No 6.56:
Question 20:
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
(a) −1
(b) 1
(c) 0
(d) None of these
Answer:
Given:
Now,
Hence, the correct option is (b).
Page No 6.56:
Question 21:
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
(a) m2 − n2
(b) m2n2
(c) n2 − m2
(d) m2 + n2
Answer:
Given:
Squaring and adding these equations, we have
Hence, the correct option is (d).
Page No 6.57:
Question 22:
If cos A + cos2 A = 1, then sin2 A + sin4 A =
(a) −1
(b) 0
(c) 1
(d) None of these
Answer:
Given:
So,
Hence, the correct option is (c).
Page No 6.57:
Question 23:
If x = a sec θ cos Ï, y = b sec θ sin Ï and z = c tan θ, then =
(a)
(b)
(c)
(d)
Answer:
Given:
Now,
Hence, the correct option is (d).
Page No 6.57:
Question 24:
If a cos θ − b sin θ = c, then a sin θ + b cos θ =
(a)
(b)
(c)
(d) None of these
Answer:
Given:
Squaring on both sides, we have
Hence, the correct option is (b).
Page No 6.57:
Question 25:
9 sec2 A − 9 tan2 A is equal to
(a) 1
(b) 9
(c) 8
(d) 0
Answer:
Given:
We know that,
Therefore,
Hence, the correct option is (b).
Page No 6.57:
Question 26:
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) =
(a) 0
(b) 1
(c) 1
(d) −1
Answer:
The given expression is .
Simplifying the given expression, we have
Disclaimer: None of the given options match with the answer.
Page No 6.57:
Question 27:
(sec A + tan A) (1 − sin A) =
(a) sec A
(b) sin A
(c) cosec A
(d) cos A
Answer:
The given expression is .
Simplifying the given expression, we have
Therefore, the correct option is (d).
Page No 6.57:
Question 28:
is equal to
(a) sec2A
(b) −1
(c) cot2A
(d) tan2A
Answer:
Given:
Therefore, the correct option is (d).
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