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Page No 273:

Question 1:

If sin θ=32, find the value of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that sin θ = perpendicularhypotenuse= ABAC = 32 .

So, if AB = 3k, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ BC2 = AC2 - AB2 = (2k)2 - (3k)2
⇒ BC2 = 4k2 - 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
   cos θ  = BCAC = k2k = 12
   tan θ  = ABBC = 3kk = 3

 ∴ cot θ  = 1tan θ = 13, cosec θ = 1sin θ = 23 and sec θ  = 1cos θ = 2

Page No 273:

Question 2:

If cos θ=725  find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ .
Now, we know that cos θ = Basehypotenuse = BCAC  = 725 .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ AB2 = AC2 - BC2 = (25k)2 - (7k)2.
⇒ AB2 = 625k2 - 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
   sin θ = ABAC  = 24k25k = 2425 
   tan θ = ABBC = 24k7k = 247 
 ∴ cot θ = 1tan θ = 724 , cosec θ = 1sin θ = 2524  and sec θ  = 1cos θ = 257 

Page No 273:

Question 3:

If tan θ=158 find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that tan θ = PerpendicularBase = ABBC = 158.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
  sin θ  = ABAC = 15k17k = 1517
  cos θ  = BCAC = 8k17k = 817

∴ cot θ  = 1tan θ = 815, cosec θ = 1sin θ = 1715 and sec θ  = 1cos θ = 178

Page No 273:

Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that cot θbasePerpendicular = BCAB = 2.


So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = 5k
Now, finding the other T-ratios using their definitions, we get:
   sin θ  = ABAC = k5k = 15
   cos θ  = BCAC = 2k5k = 25

∴ tan θ  = 1cot θ = 12, cosec θ = 1sin θ = 5 and sec θ  = 1cos θ = 52

Page No 273:

Question 5:

If cosec θ = 10, the find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that cosec θ = HypotenusePerpendicular = ACAB= 101.

So, if AC = (10)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ BC2 = AC2 - AB2 = 10k2 - k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
   tan θ  = ABBC = k3k = 13

   cos θ  = BCAC = 3k10k = 310

 ∴ sin θ=1cosec θ=110, cot θ  = 1tan θ = 3 and sec θ  = 1cos θ = 103

Page No 273:

Question 6:

If cot θ = 158, show then evaluate (2+2 sinθ)(1-sinθ)(1+cosθ)(2-2cosθ).

Answer:

Let us consider a right ABC, right angled at B.
Now, we know that cot θ =BasePerpendicular =BCAB = 158.

Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (8k)2 + (15k)2
⇒ AC2 = 64k2 + 225k2 = 289k2
⇒ AC = 17k

Finding out the value of sin θ and cos θ using their definitions, we have:
sin θ = ABAC = 8k17k = 817
cos θ = BCAC = 15k17k = 1517

Substituting these values in the given expression, we get:
(2 + 2sin θ)(1 -sin θ)(1 + cos θ)(2 -2cos θ)=(2 + 2×817)(1 - 817)(1 + 1517)(2 - 2×1517)=(2 + 1617)(1-817)(1+1517)(2 -3017)=(34 + 1617)(917)(3217)(34 - 3017)=5017×9173217×417=22564

Page No 273:

Question 7:

If tan θ = 43, show that (sin θ + cos θ) = 75.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now, we know that  tan θABBC43.

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2 + 9k2 = 25k2
⇒ AC = 5k
Finding out the values of sin θ and cos θ using their definitions, we have:
sin θ = ABAC = 4k5k = 45
cos θ = BCAC = 3k5k = 35
Substituting these values in the given expression, we get:
(sin θ + cos θ) = (45 + 35 ) = (75) = RHS
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 8:

If tan θ = 17, show that cosec2θ-sec2θcosec2θ+sec2θ=34.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now it is given that tan θABBC17.

So, if AB = k, then BC = 7k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + (7k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 22k
Now, finding out the values of the other trigonometric ratios, we have:
sin θ  = ABAC = k22k = 122
cos θ  = BCAC = 7 k22k = 722
∴ cosec θ  = 1sin θ = 22 and sec θ   = 1cos θ = 227
Substituting the values of cosec θ  and sec θ  in the given expression, we get:
 cosec2θ - sec2θcosec2θ + sec2θ=(22)2 - 2272(22)2 + 2272=8 - 878 + 87=56 - 8756 + 87=4864 = 34 = RHS
 i.e., LHS = RHS
 
Hence proved.

Page No 273:

Question 9:

If cosec θ = 2, show that cotθ+sinθ1+cosθ=2.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now, it is given that cosec θ = 2.
Also, sin θ  = 1cosecθ = 12 = ABAC


So, if AB = k, then AC = 2k, where k is a positive number.  
Using Pythagoras theorem, we have:
⇒ AC2 = AB2  + BC2
⇒ BC2 = AC2 - AB2
⇒ BC2 = (2k)2 - (k)2
⇒ BC2 = 3k2
⇒ BC = 3k
Finding out the other T-ratios using their definitions, we get:
cos θ = BCAC = 3k2k = 32
tan θ = ABBC = k3k = 13
cot θ = 1tanθ = 3
Substituting these values in the given expression, we get:
cot θ + sinθ1 + cosθ= 3 + 121 + 32= 3 + 122+ 32

=3+12+3=32+3+12+3=23+3+12+3=22+32+3=2
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 10:

If sec θ = 54, verify that tanθ1+tan2θ=sinθsecθ.

Answer:

Let us consider a right ABC, right angled at B and C=θ.
Now, it is given that sec θ = 54.
Also, cos θ  = 1secθ = 45 = BCAC
 
So, if BC = 4k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2  + BC2
⇒ AB2 = AC2 - BC2
⇒ AB2 = (5k)2 - (4k)2 = 25k2 - 16k2
⇒ AB2 = 9k2
⇒ AB = 3k

Finding out the other T-ratios using their definitions, we get:
   sin θ = ABAC = 3k5k = 35

   tan θ = ABBC = 3k4k = 34

Substituting the values in LHS and RHS, we get:
   LHS=tan θ1 + tan2θ =341 + 342=341 + 916=3416 + 916=1225 RHS=sinθsecθ = 3554 = 1225
  i.e., LHS = RHS
 
Hence proved.

Page No 273:

Question 11:

If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.

Answer:

Let us consider a right ABC right angled at B.
Now, we know that cos θ = 0.6 = BCAC35

 
 So, if BC = 3k, then AC = 5k, where k is a positive number.
 Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AB2 = AC2 - BC2
⇒ AB2 = (5k)2 - (3k)2 = 25k2 - 9k2
⇒ AB2 = 16k2
⇒ AB = 4k

Finding out the other T-ratios using their definitions, we get:
   sin θ = ABAC = 4k5k = 45

   tan θ = ABBC = 4k3k = 43

Substituting the values in the given expression, we get:
 5 sin θ - 3 tan θ
545 - 343 4 - 4 = 0 = RHS
 i.e., LHS  = RHS
 
Hence proved.

Page No 273:

Question 12:

If 3 tan θ = 4, show that 4cosθ-sinθ2cosθ+sinθ=45.

Answer:

Let us consider a right ABC right angled at B and C=θ.
We know that tan θ = ABBC43

So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2  + BC2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 5k

Now, we have:

sin θ = ABAC = 4k5k = 45

cos θ = BCAC = 3k5k = 35

Substituting these values in the given expression, we get:

 4 cosθ - sinθ2 cosθ + sinθ= 435 - 45235 + 45=125-4565+45= 12 - 456 + 45= 810= 45 = RHS
i.e., LHS = RHS

Hence proved.

Page No 273:

Question 13:

If tan θ = ab, show that a sinθ-b cosθa sinθ+b cosθ=a2-b2a2+b2.

Answer:

It is given that tan θ = ab.

LHS = a sinθ - b cosθa sinθ + b cosθ
 Dividing the numerator and denominator by cos θ, we get:

 a tan θ - ba tan θ + b       (∵ tan θ = sin θcos θ)
Now, substituting the value of tan θ in the above expression, we get:
 aab - baab + b= a2b - ba2b + b= a2 - b2a2 + b2 = RHS
  i.e., LHS = RHS

 Hence proved.

Page No 273:

Question 14:

If 3 cot θ = 2, show that 4sinθ-3cosθ2sinθ+6cosθ=13.

Answer:

It is given that cot θ = 23.

LHS  = 4 sinθ - 3 cosθ2 sinθ + 6 cosθ
Dividing the above expression by sin θ, we get:
4 - 3 cot θ2 + 6 cot θ                     [∵ cot θ = cosθsinθ]
Now, substituting the values of cot θ in the above expression, we get:
 4 - 3232 + 623= 4 - 22 + 4 = 26=13
 i.e., LHS = RHS
 
Hence proved.

Page No 273:

Question 15:

If sec θ = 178, verify that 3-4sin2θ4cos2θ-3=3-tan2θ1-3tan2θ.

Answer:

It is given that sec θ = 178.

Let us consider a right ABC right angled at B and C=θ.
We know that cos θ = 1sec θ= 817 = BCAC
 
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2 = (17k)2 - (8k)2
⇒ AB2 = 289k2 - 64k2 = 225k2
⇒ AB = 15k.

Now, tan θ  = ABBC = 158 and sin θ = ABAC = 15k17k= 1517

The given expression is 3 - 4sin2θ4cos2θ- 3 = 3 - tan2θ1 - 3tan2θ.
 
 Substituting the values in the above expression, we get:
 LHS= 3 - 41517248172 - 3 = 3 - 900289256289- 3 = 867-900256-867= -33-611=33611

RHS = 3-15821-31582=3-225641-67564=192-22564-675=-33-611=33611

∴ LHS = RHS
Hence proved.

Page No 273:

Question 16:

If tan θ = 2021, show that1-sinθ+cosθ1+sinθ+cosθ=37.

Answer:

Let us consider a right ABC right angled at B and C=θ.
Now, we know that tan θABBC = 2021

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
 AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒  AC = 29k
Now, sin θ = ABAC = 2029 and cos θ = BCAC = 2129

Substituting these values in the given expression, we get:
  LHS=1 - sinθ + cosθ1 + sinθ + cosθ= 1 - 2029 + 21291 + 2029 + 2129= 29 - 20 + 212929 + 20 + 2129= 3070 = 37 = RHS
∴ LHS = RHS

Hence proved.

Page No 273:

Question 17:

In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.

Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.

Answer:


Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = 122  + 52 = 144 + 25
⇒ AC2 = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A = ABAC= 1213
(ii) cosec A = 1sin A = ACBC = 135

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C = BCAC = 513
(iv) cosec C = 1sin C = ACAB =1312



Page No 274:

Question 18:

In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure

Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C

Answer:


Using Pythagoras theorem, we get:
 AC2 = AB2 + BC2
⇒ AC2= (24)2 + (7)2
⇒ AC2 = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i)  sin A = BCAC = 725

(ii) cos A = ABAC = 2425

Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C = ABAC= 2425

(iv) cos C = BCAC = 725

Page No 274:

Question 19:

Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos2θ − sin2θ) = 41841.

Answer:


Using Pythagoras theorem, we get:
AB2  =  AC2  + BC2
⇒ AC2 = AB2 - BC2
⇒ AC2 = (29)2 - (21)2
⇒ AC2 = 841- 441
⇒ AC2 = 400
⇒ AC = 400 = 20 units

Now, sin θ = ACAB = 2029 and cos θ = BCAB = 2129

 cos2 θ - sin2 θ = 21292 - 20292 =441841-400841=41841

Hence Proved.



Page No 276:

Question 1:

If cosθ = 45, then tanθ = ?
(a) 34
(b) 43
(c) 35
(d) 53

Answer:

(a) 34
Since cosθ = 45 but cosθ = ABAC
So, ABAC = 45
Thus, AB = 4k and AC = 5k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 - AB2
⇒ BC2  = (5k)2 - (4k)2
⇒ BC2 = 9k2
⇒ BC = 3k
∴ tanθ = BCAB = 34

Page No 276:

Question 2:

If sinθ = 12, then cotθ = ?
(a) 13
(b) 3
(c) 32
(d) 1

Answer:

(b) 3

Given: sinθ = 12, but sinθ = BCAC
So, BCAC = 12
Thus, BC = k and AC = 2k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 - BC2
AB2 = (2k)2 - (k)2
AB2 = 3k2
AB = 3k
So, tanθ = BCAB = k3k = 13
∴ cotθ = 1tanθ = 3

Page No 276:

Question 3:

If tan θ = 34, then cosec θ = ?
(a) 43
(b) 45
(c) 54
(d) 53

Answer:

(d) 53

Given: tan θ = 34, but tan θ = PerpendicularBase
So, let us first draw a right ABC right angled at B and A=θ.
So, BCAB = 34
Thus, BC = 3k and AB = 4k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (4k)2 + (3k)2
⇒ AC2 = 25k2
⇒ AC = 5k
∴ cosec θ = ACBC = 5k3k = 53

Page No 276:

Question 4:

If cosec θ = 10, then sec θ = ?
(a) 310
(b) 103
(c) 110
(d) 210

Answer:

(b) 103
Let us first draw a right ABC right angled at B and A=θ.
Given: cosec θ10, but sin θ1cosec θ = 110
Also, sin θ = PerpendicularHypotenuse = BCAC
So, BCAC = 110
Thus, BC = k and AC = 10 k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 - BC2
AB2 = (10 k)2 - (k)2
AB2 = 9k2
AB = 3k
∴ sec θ = ACAB = 10k3k=103

Page No 276:

Question 5:

If tan θ = 815, then cosec θ = ?
(a) 178
(b) 817
(c) 1715
(d) 1517

Answer:

(a) 178

Let us first draw a right ABC right angled at B and A=θ.
Given: tan θ = 815, but tan θ = BCAB
So, BCAB = 815
Thus, BC = 8k and AB = 15k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (15k)2 + (8k)2
⇒ AC2 = 289k2
⇒ AC = 17k
∴ cosec θ = ACBC=17k8k=178

Page No 276:

Question 6:

If sin θ ab, then cos θ = ?
(a) bb2-a2
(b) b2-a2b
(c) ab2-a2
(d) ba

Answer:

(b) b2-a2b
Let us first draw a right ABC right angled at B and A=θ.
Given: sin θ = ab, but sin θ = BCAC
So, BCAC = ab
Thus, BC = ak and AC = bk


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2
⇒ AB2 = (bk)2 - (ak)2
⇒ AB2 = b2-a2k2
⇒ AB = (b2-a2)k
∴ cos θ  = ABAC = b2-a2kbk = b2-a2b

Page No 276:

Question 7:

If tan θ = 3, then sec θ = ?
(a) 23
(b) 32
(c) 12
(d) 2

Answer:

(d) 2

Let us first draw a right ABC right angled at B and A=θ.
Given: tan θ = 3
But tan θ = BCAB
So, BCAB = 31
Thus, BC = 3k and AB = k


Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = (3 k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ = ACAB = 2kk = 21

Page No 276:

Question 8:

If sec θ = 257, then sin θ = ?
(a) 724
(b) 247
(c) 2425
(d) None of these

Answer:

(c) 2425
Let us first draw a right ABC right angled at B and A=θ.
Given: sec θ = 257
But cos θ1sec θABAC = 725
Thus, AC = 25k and AB = 7k


Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
BC2 = AC2 -AB2
BC2 = (25k)2 - (7k)2
BC2= 576k2
BC = 24k

∴ sin θ = BCAC = 24k25k = 2425

Page No 276:

Question 9:

If sin θ = 32, then (cosecθ + cotθ) = ?
(a) 23
(b) 233
(c) (2+3)
(d) 3

Answer:

(d) 3
Let us first draw a right ABC right angled at B and A=θ.
sin θ = 32 = BCAC
So, BC = 3k and AC = 2k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AB2 = AC2 - BC2
AB2 = (2k)2  - (3k)2
AB2 = k2
AB  = k

So, cot θ = ABBC = k3k = 13
cosec θ = 1sin θ = 23
∴ (cosec θ + cot θ) = 23 + 13 = 33 = 3



Page No 277:

Question 10:

If tan θ = 43, then (sinθ + cosθ) = ?
(a) 73
(b) 74
(c) 75
(d) 57

Answer:

(c) 75
Let us first draw a right ABC right angled at B and A=θ.
tan θ = 43 = BCAB
So, AB = 3k and BC = 4k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AC2 = (3k)2 + (4k)2
AC2 = 25k2
AC= 5k
Thus, sin θ = BCAC = 45
and cos θ =ABAC = 35
∴ (sin θ + cos θ) = ( 45 + 35) = 75

Page No 277:

Question 11:

If 3x = cosec θ and 3x = cot θ, than 3x2-1x2=?
(a) 127
(b) 181
(c) 13
(d) 19

Answer:

(c) 13
Given: 3x = cosec θ and 3x = cot θ
Also, we can deduce that x = cosec θ3 and 1x = cot θ3.
So, substituting the values of x and 1x in the given expression, we get:
3x2-1x2 = 3cosec θ32 - cot θ32
= 3cosec2θ9 - cot2θ9
= 39cosec2θ -cot2θ
= 13       [By using the identity: cosec2θ- cot2θ = 1]

Page No 277:

Question 12:

If 2x = sec A and 2x = tan A, then 2x2-1x2=? =?
(a) 12
(b) 14
(c) 18
(d) 116

Answer:

(a) 12
Given: 2x = sec A and 2x = tan A
Also, we can deduce that x = sec A2 and 1x = tan A2.
So, substituting the values of x and 1x in the given expression, we get:
2x2-1x2 = 2sec A22 - tan A22
= 2sec2A4 - tan2A4
= 24sec2A -tan2A
= 12                  [By using the identity: sec2θ- tan2θ = 1]

Page No 277:

Question 13:

If (tan θ + cot θ) = 5, then (tan2 θ + cot2 θ) = ?

(a) 27
(b) 25
(c) 24
(d) 23

Answer:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25       [∵ tan θ = 1cot θ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23

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Question 14:

If (cos θ + sec θ) = 52, then (cos2 θ + sec2 θ) = ?
(a) 214
(b) 174
(c) 294
(d) 334

Answer:

(b) 174
We have (cos θ +sec θ) = 52
Squaring both sides, we get:
(cos θ + sec θ)2 = (52)2
cos2 θ + sec2 θ + 2 cos θ sec θ = 254
cos2 θ + sec2 θ + 2 = 254      [∵ sec θ = 1cos θ]
cos2 θ + sec2 θ = 254 − 2 = 174

Page No 277:

Question 15:

If tan θ = 17, then (cosec2θ-sec2θ)(cosec2θ+sec2θ)=? = ?
(a) -23
(b) -34
(c) 23
(d) 34

Answer:

(d) 34
=cosec2θ - sec2θcosec2θ + sec2θ = sin2θ1sin2θ-1cos2θsin2θ1sin2θ +1cos2θ      [Multiplying  the numerator and denominator by sin2θ]= 1 - tan2θ1 + tan2θ
= 1-171 +17 = 68 = 34

Page No 277:

Question 16:

If 7 tan θ = 4, then (7sinθ-3cosθ)(7sinθ+3cosθ) = ?
(a) 17
(b) 57
(c) 37
(d) 514

Answer:

(a) 17

7 tan θ = 4

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
1cosθ7sinθ - 3cosθ1cosθ7sinθ + 3cosθ

= 7tanθ - 37tanθ + 3

= 4 - 34 + 3         [∵ 7 tan θ = 4]
17

Page No 277:

Question 17:

If 3 cot θ = 4, then (5sinθ+3cosθ)(5sinθ-3cosθ) = ?
(a) 13
(b) 3
(c) 19
(d) 9

Answer:

(d) 9

We have 5sinθ + 3cosθ5sinθ - 3cosθ.

Dividing the numerator and denominator of the given expression by sin θ, we get:
 1sinθ5sinθ + 3cosθ1sinθ5sinθ - 3cosθ

= 5 + 3cot θ5 - 3cot θ

= 5 + 45 - 4 = 9              [∵ 3 cot θ = 4]

Page No 277:

Question 18:

If θ = ab, then (a sinθ-b cosθ)(a sinθ+b cosθ) = ?
(a) (a2+b2)(a2-b2)
(b) (a2-b2)(a2+b2)
(c) a2(a2+b2)
(d) b2(a2+b2)

Answer:

(b) a2-b2a2 +b2
We have tan θ = ab

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
 asin θ - bcos θasin θ + bcos θ=1cosθasin θ - bcos θ1cosθasin θ + bcos θ =atanθ - b atanθ + b =a2b-ba2b+b =a2 -b2a2 +b2



Page No 278:

Question 19:

If cot θ = , then (2+2sinθ)(1-sinθ)(1+cosθ)(2-2cosθ) = ?
(a) 64225
(b) 22564
(c) 64289
(d) 28964

Answer:

(b) 22564
We have:
 
2+2sinθ1-sinθ1+cosθ2-2cosθ=21+sinθ1-sinθ21+cosθ1-cosθ= 21-sin2θ21-cos2θ= cos2θsin2θ= cot2θ =22564

Page No 278:

Question 20:

In the given ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm. Then, cos A = ?
Figure

(a) 724
(b) 725
(c) 2425
(d) None of thse

Answer:

(c) 2425


Here, AB = 24 cm and BC = 7 cm
Now, using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AC2 = (24)2 + (7)2
AC2 = 625
AC = 25 cm
∴ cos A = ABAC= 2425



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