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#### Question 27:

$3\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{9}{3}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(9{x}^{2}-\frac{9}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{\left(3x\right)}^{2}-{\left(\frac{3}{x}\right)}^{2}\right]$
$=\frac{1}{3}\left[{\left(\mathrm{cosec}\theta \right)}^{2}-{\left(\mathrm{cot}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left({\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

#### Question 28:

$3\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{9}{3}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(9{x}^{2}-\frac{9}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{\left(3x\right)}^{2}-{\left(\frac{3}{x}\right)}^{2}\right]$
$=\frac{1}{3}\left[{\left(\mathrm{cosec}\theta \right)}^{2}-{\left(\mathrm{cot}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left({\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

From the given right-angled triangle, we have:

#### Question 29:

From the given right-angled triangle, we have:

From the given right-angled triangle, we have:

#### Question 30:

From the given right-angled triangle, we have:

From right-angled âˆ†ABC, we have:
â€‹

#### Question 1:

From right-angled âˆ†ABC, we have:
â€‹

On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

#### Question 2:

On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

Ans

#### Question 3:

Ans

On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
$=\left(\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

#### Question 4:

On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
$=\left(\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o  sin 30o

#### Question 5:

On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o  sin 30o

As we know that,

#### Question 6:

As we know that,

On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

#### Question 7:

On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

As we know that,
â€‹

#### Question 8:

As we know that,
â€‹

On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − $\frac{3}{4}$ sec2 45o + $\frac{1}{4}$ cosec2 30o

#### Question 9:

On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − $\frac{3}{4}$ sec2 45o + $\frac{1}{4}$ cosec2 30o

As we know that,
â€‹

As we know that,
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(i)

Hence, LHS = RHS

(ii)

Hence, LHS = RHS

1sin60°cos60°

#### Question 11:

(i)

Hence, LHS = RHS

(ii)

Hence, LHS = RHS

1sin60°cos60°

(i) sin 60o cos 30o − cos 60o sin 30o

∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

â€‹
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o

∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
$=2×\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}=1$
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

#### Question 12:

(i) sin 60o cos 30o − cos 60o sin 30o

∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

â€‹
∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o

∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
$=2×\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}=1$
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

A = 45o
⇒ 2A = 2 $×$ 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2A − 1 = 2 cos2 45o − 1 =
Now, 1 − 2 sin2A
∴ cos 2A 2 cos2A − 1 = 1 − 2 sin2A

#### Question 13:

A = 45o
⇒ 2A = 2 $×$ 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2A − 1 = 2 cos2 45o − 1 =
Now, 1 − 2 sin2A
∴ cos 2A 2 cos2A − 1 = 1 − 2 sin2A

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

â€‹(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

=2tanA1+tan2A

#### Question 14:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

â€‹(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

=2tanA1+tan2A

A = 60o and B = 30o
Now, A + B = 60o + 30oâ€‹ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o
∴â€‹ cos (A + B) = cos A cos B − sin A sin B

#### Question 15:

A = 60o and B = 30o
Now, A + B = 60o + 30oâ€‹ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o
∴â€‹ cos (A + B) = cos A cos B − sin A sin B

(i) sin (A − B) = sin 30o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o

∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴â€‹ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = $\frac{1}{\sqrt{3}}$

∴â€‹ tan (AB) =

#### Question 16:

(i) sin (A − B) = sin 30o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o

∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴â€‹ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = $\frac{1}{\sqrt{3}}$

∴â€‹ tan (AB) =

Given:

13

#### Question 17:

Given:

13

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ tan 60o = $\sqrt{3}$

#### Question 18:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ tan 60o = $\sqrt{3}$

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ cos 30o = $\frac{\sqrt{3}}{2}$

#### Question 19:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ cos 30o = $\frac{\sqrt{3}}{2}$

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ sin 30o = $\frac{1}{2}$

#### Question 20:

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ sin 30o = $\frac{1}{2}$

Disclaimer: $\mathrm{cos}15°$ can also be calculated by taking .

#### Question 21:

Disclaimer: $\mathrm{cos}15°$ can also be calculated by taking .

As we know that,
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As we know that,
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As we know that,

#### Question 25:

As we know that,

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A B) = tan 30o       [âˆµ tan 30o = $\frac{1}{\sqrt{3}}$]
A − B = 30o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒â€‹ tan (A + B) =  tan 60o        [âˆµ tan 60o = $\sqrt{3}$]
A + B = 60o                           ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

#### Question 26:

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A B) = tan 30o       [âˆµ tan 30o = $\frac{1}{\sqrt{3}}$]
A − B = 30o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒â€‹ tan (A + B) =  tan 60o        [âˆµ tan 60o = $\sqrt{3}$]
A + B = 60o                           ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

As we know that,

#### Question 31:

As we know that,

(i) As we know that,
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(ii) As we know that,
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#### Question 1:

(i) As we know that,
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(ii) As we know that,
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#### Question 8:

As we know that,
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Given: tan= 3cotx

â€‹

#### Question 9:

Given: tan= 3cotx

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As we know that,
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