Direction: Study the following questions carefully and choose the right answer:
Important for :
1
For any real values of
 sec Θ – 1 sec Θ + 1
= ?
» Explain it
C
 sec Θ – 1 sec Θ + 1
=
 sec Θ – 1 sec Θ + 1
×
 sec Θ – 1 sec Θ – 1

[Rationalising the numerator and the denominator]

 (sec Θ – 1)2 sec2 Θ – 1

 (sec Θ – 1)2 tan2 Θ

[∵ sec2 Θ – 1 = tan2 Θ]

 = sec Θ – 1 tan Θ

 = sec Θ – 1 tan Θ tan Θ

 1 cos Θ
=
 1 tan Θ
 sin Θ cos Θ

[ ∵ sec Θ =
 1 cos Θ
& tan Θ =
 sin Θ cos Θ
]

 = 1 – 1 sin Θ tan Θ

= cosec Θ – cot Θ

 [ ∵ 1 = cosec Θ & 1 = cot Θ ] sin Θ tan Θ

Hence, option C is correct.

2
In a ΔABC, ∠B =  π/3 , ∠C =  π/4  and D divides BC internally in the ratio 1 : 3 then (sin ∠BAD) / (sin ∠CAD) is equal to
» Explain it
C
 Given that ∠B = π and ∠C = π 3 4
 and BD = 1 CD 3

In ΔABD,

 3 /2

 3
. BD        ...(i)

In ΔACD,

 1/ 2

Equation (i) ÷ (ii),

=
 3
. BD

 3
. BD  =
 3
×  1      [ ∵  BD  =  1 ]
2 CD 2 3 CD 3

 = 1 6

Hence, option C is correct.

3
If sin 3A = cos (A – 26°), where 3A is an acute angle then the value of A is
» Explain it
A
sin 3A = cos (A – 26°)

⇒  cos (90° – 3A) = cos (A – 26°)

[∵ cos (90° – Θ) = sin Θ]

⇒  90° – 3A = A – 26°

⇒  4A = 116°

⇒  A = 29°

Hence, option A is correct.

4
 Value of sec2 Θ – sin2 Θ – 2 sin4 Θ is 2 cos4 Θ – cos2 Θ
» Explain it
A
 sec2 Θ – sin2 Θ – 2 sin4 Θ 2 cos4 Θ – cos2 Θ

 = sec2 Θ – sin2 Θ (1 – 2 sin2 Θ) cos2 Θ (2 cos2 Θ – 1)

 = sec2 Θ – sin2 Θ cos 2Θ cos2 Θ cos 2Θ

[∵ cos 2Θ  =  1 – 2 sin2 Θ  =  2 cos2 Θ – 1]

 = sec2 Θ – sin2 Θ cos2 Θ

= sec2 Θ – tan2 Θ

 [ ∵ sin Θ = tan Θ ] cos Θ

= 1

[∵ sec2 Θ – tan2 Θ = 1]

Hence, option A is correct.

5
If x = a (sin Θ + cos Θ), y = b (sin Θ – cos Θ)

 then the value of x2 + y2 is a2 b2
» Explain it
C
Given that x = a (sin Θ + cos Θ)   and   y = b (sin Θ – cos Θ)

 x2 + y2 a2 b2

 = a2 (sin Θ + cos Θ)2 + b2 (sin Θ – cos Θ)2 a2 b2

= (sin Θ + cos Θ)2 + (sin Θ – cos Θ)2

= sin2 Θ + cos2 Θ + 2 sin Θ cos Θ + sin2 Θ + cos2 Θ – 2 sin Θ cos Θ

= 2(sin2 Θ + cos2 Θ)

= 2 × 1

[∵ sin2 Θ + cos2 Θ = 1]

= 2

Hence, option C is correct.