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Question-1

In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) SinA , CosA
(ii) SinC , CosC.

Solution:

AB = 24 cm
BC = 7 cm

Question-2

In the figure, find tan P – cot R.

Solution:
In a right angled Δ PQR
PR2 = PQ2 + QR2
132 = 122 + QR2
QR2 = 169 – 144 = 25
QR = 5

tan P - cot R

tanP-cotP = 0

Question-3

sin A =, calculate cos A and tan A.

Solution:
Let the right angled ΔABC be right angled at B.

In a right angled Δ ABC

AC2 = AB2 + BC2
16 = AB2 + BC
AB2 = 16 – 9 = 7
AB =
sin A =
cos A = =

tan A = .

Question-4

Given 15 cot A = 8, find sin A and sec A.

Solution:
15 cot A = 8
cot A =

Let ABC be the Δ right angled at B
Cot A = 8/15
AB = 8, BC = 15
AC2 = AB2 + BC2
      = 82 + 152
      = 64 + 225
      = 289
 AC = = 17
sin A = =
sec A = = =

Question-5

Given sec θ = calculate all other trigonometric Ratios.

Solution:



In a right angled Δ ABC,
AC2 = AB2 + BC2
132 - 122 = AB2
AB2 = 169 - 144
AB = 5



Question-6

If A and B are acute angles such that cos A = cos B, then show that A = B.

Solution:
In a right ΔABC, 

Since opposite sides of equal angles are equal.
 = 450.

Question-7

If cot θ = , evaluate : (i) (ii) cot2θ.

Solution:
If cot θ = , Sin θ =
(i) =
                                   =  
                                   
 
      (ii)

Question-8

If 3 cot A = 4, Check whether = cos2 A – sin2 A or not.

Solution:
If 3 cot A = 4
= = = ………………… (1)
cos2 A – sin2 A = sin2 A(cot2 A - 1) Dividing and multiplying by sin2 A
                   = sin2 A
                   = sin2 A= sin2 A
                   = = …………………………. (2)
Since (1) = (2)
LHS = RHS.

Question-9

In triangle ABC, right-angled at B, if tan A =find the value of (i) sin A cos C +cos A sin C (ii)cos A cos C – sin A sin C.

Solution:
If tan A =
In a right angled Δ ABC, right angled at B
AC2 = AB2 + BC2
AC = = 2
sin A = , cos A = sin C = cos C = 1/2

(i) sin A cos C + cos A sin C
= = = 1

(ii) cos A cos C – sin A sin C
= 0.

Question-10

In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution:

PR + QR = 25
PQ = 5 cm
In a right angled Δ PQR, PQ2 + QR2 = PR2 …………………… (1)
Let PR = x
QR = (25 – x)
From (1) x2 = 52 + (25 – x)2
x2 = 25 + (25)2 – 2 × 25x + x2
    = 625 + 25 – 50x
50x = 650
x = = 13
PR = 13 cm
QR = 12 cm
PQ = 5 cm
sin P = =
cos P = =
tan P = = .

Question-11

State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ =
.

Solution:
(i) False. The value of tan 90is greater than 1

(ii) True. sec A = cos A = as 12 the hypotenuse is the largest side of triangle.

(iii) False. cos A is the abbreviation for cosine of angle A

(iv) No, False. cos A is not the product of cot and A

(v) False, Because the hypotenuse is the longest side whereas in sin θ = , 3 which is the denominator is cannot be the hypothenuse.
 

Question-12

Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60° = 2
(iii)
(iv)

(v) .

Solution:

cbse_10_math_images8_GSG/5ttr.gif





=   

 

Question-13

Choose the correct option and justify your choice:
(i) =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°

(ii) =

(A) tan 90° (B) 1 (C) sin 45° (D) 0

(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°

(iv)=

(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°.

Solution:
(i) = =
                      =
                      = =
                      = sin 600
Answer: A

(ii) 
= = 0

Answer: D
(iii) sin 2A = 2 sin A is true when A = 00


Answer: A
(iv) = =
                     =
                     =
                     =
                     =×
                     = = = tan 600.

Answer: C

Question-14

tan (A + B) = and tan (A – B) =; 0° < A + B 90°; A > B, find A and B.

Solution:
If tan(A + B) =
tan(A - B) = 1/
If tan A+B =
A + B = 600 ………………………(1)
tan (A – B) =
A – B = 300 ……………………..(2)
Adding (1) and
 (2)2A = 900
A = 450
 B = 150 .

Question-15

State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin increases as increases.
(iii) The value of cos increases as increases.
(iv) sin = cos for all values of .
(v) cot A is not defined for A = 0°.

Solution:
(i) sin(A + B) = sin A + sin B
False, suppose A = 300, B = 600 because sin(A + B) = sin 900 = 1

sin A + sin B = 1/2 + √3/2 ≠ 1

(ii) Value of sin q as θ increase
True

(iii) Value of cos θ as θ increases
False

(iv) sinθ = cosθ for all values of θ
False

(v) cot A is not defined for A = 00
True.

Question-16

Evaluate:
(i)
(ii) 

(iii) cos480 – sin 420
(iv)cosec 310 – sec 590.

Solution:
(i)
    =
    = = 1

(ii) =  = 1

(iii) cos480 – sin 420
cos480 = cos(90 - 420)
          = sin 420
cos 480 – sin 420 = sin 420 – sin 420
                           = 0

(iv) cosec 310 – sec 590
cosec 310 = cosec(900 - 590)
              = sec 590
LHS = sec 590 – sec 590 = 0.

Question-17

Show that tan 480 tan 230 tan 420 tan 670 = 1.

Solution:
(i) tan 480.tan230. tan 420 tan670 = 1
LHS = tan 480. tan 420 tan 230 tan670
       
= tan 480. tan(900 - 480). tan(900 - 670)  tan 670
        = tan 480. cot 480. cot 670 tan 670.
        = 1 × 1 = 1.

Question-18

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:
tan 2A = cot (A – 18) ………………. (1)
where 2A is an acute angle
tan 2A = cot(90 – 2A) = cot(A – 18)
90 – 2A and A - 18 are acute angles.
90 – 2A = A – 18
3A = 108
A = 360.

Question-19

If tan A = cot B, prove that A + B = 90°.

Solution:
If tan A = cot B …………… (1)
tan A = cot (90 – A)
From (1)
cot (90 - A) = cot B
    90 – A = B
or A + B = 900.

Question-20

If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:
If sec 4A = cosec (A – 200)
sec 4A = cosec (90 – 4A)
cosec(90 – 4A) = cosec (A – 200)
Since (90 – 4A) and A – 200 are acute angles
90 – 4A = A – 200
90 + 20 = 5A
         A =
            = 220.

Question-21

A, B and C are interior angles of a triangle ABC, then show that sin = cos .

Solution:


Question-22

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

Question-23

Express the trigonometric ratios sin A, cos A and tan A in terms of sec A.

Solution:

Question-24

Write all the other trigonometric ratios of A in terms of sec A.

Solution:
sin A = 
           =
cos A =
tan A = = ÷
        = × sec A.

            =

Question-25

Evaluate:
(i) (ii) sin 250. cos 650 + cos 250 sin 650.

Solution:

Question-26

Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =
(A) 1 (B) 9 (C) 8 (D) 0

(ii) (1 + tan θ + secθ ) (1 + cot θ – cosec θ ) =
(A) 0 (B) 1 (C) 2 (D) –1

(iii) (sec A + tan A) (1 – sin A) =
(A)sec A (B) sin A (C) cosec A (D) cos A

(iv)
(A) sec2 A (B) –1 (C) cot2 A (D) tan2 A
.

Solution:
(i) Answer (B)
9 sec2A - 9 tan2A
   = 9 (1 + tan2A) – 9 tan2A
   = 9 + 9 tan2A - 9 tan2A
   = 9


(ii)Answer (C)

 
(iii) Answer (D)
(sec A + tan A) (1 – sin A)
sec A – sin A sec A + tan A – tan A sin A
+ - .sinA



(iv) Answer (D)
= = = tan2A.

Question-27

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosecθ - cot θ )2 =

(ii)= 2 secA

(iii)= 1 + secθ cosecθ

(iv) =

(v)  = Cosec A + cot A

Solution:
(i) (cosecθ - cot q )2 =
RHS =
                 
                  =
                  = cosec2θ - 2cosecθ.cotθ + cot2θ
                  = (cosec θ - cotθ)2
Hence Proved.


(ii)
                  = =
                  = =
                  = = 2 sec A

(iii)


(iv) =
LHS

RHS =
                     = 1 + cos A = LHS

(v) 
 
Using the identity cosec2A = 1 + cot2A.
Taking the LHS of the given equation

Dividing the numerator and denominator by sin A

= (Changing = cot A and
=      (
∵  using the identity cosec2A - cot2A = 1 )

(Replacing cosec2A - cot2A = (cosec A + cot A) (cosec A - cot A)  

=
=
= cosec A + cotA.

 

Question-28

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) = sec A + tan A
(ii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + Cot2A
(iii) (cosec A – sin A)(sec A - cos A) =

(iv)= .

Solution:
(i)
      = Taking the conjugate of (1-sinA) + multiplying the numerator and denominator by it.
      =
      = (Since 1 - sin2A = cos2A)
      = =
      = = sec A + tan A (As = sec A, = tan A)

(ii) (sin A + cosec A)2 + (cos A + sec A)2
      
= sin2A + cosec2A + 2sinA.cosec A + cos2A + sec2A + 2cos A. sec A
      = sin2A + cos2A + cosec2A + sec2A + 2sinA . + 2cos A.
      = 1 + 2 + 2 + cosec2A + sec2A (using cosec A = secA = )
      = 5 + 1 + tan2A + 1 + cot2A    
 (  sec2A = 1 + tan2A  and  cosec2A = 1 + cot2A)
      = 7 + tan2A + cot2A      

(iii) (cosec A - sin A) (secA - cosA)
LHS
(cosec A - sin A) (sec A - cos A)

  (since 1 - sin2A = cos2A and 1 - cos2A = sin2A)
= = sin A cosA
RHS =
                          = (Taking LCM as cosA. sinA)
                          = (since sin2A + cos2A = 1)
LHS = RHS
Hence the result.

(iv) 
.
LHS

RHS

Thus, LHS = RHS.




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