Direction: Study the following questions carefully and choose the right answer:
Important for :
1
If Θ is an acute angle and tan2Θ + 1/tan2Θ = 2 then the value of Θ is:
» Explain it
C
 tan2θ + 1 = 2 tan2θ

tan4θ + 1 = 2tan2θ

tan4θ + 1 – 2tan2θ = 0

(tan2θ)2 + (1)2 – 2 (tan2θ) (1) = 0

[ ∴   a2 + b2 – 2ab = (a + b)]

(tan2θ – 1)2 = 0 ⇒ tan2θ = 1

tanθ1 = 1  ⇒  tanθ = tan 45°   ⇒       θ = 45°.

Hence, option C is correct.

2
If α + β = 90°, then the value of (1 – sin2α) (1 – cos2α) × (1 + cot2β) (1 + tan2β) is
» Explain it
A
As per the question, α + β = 90°,    α = 90° – β.

(1 – sin2α) (1 – cos2α) × (1 + cot2β)(1 + tan2β)

[ ∵   1 + tan2β = sec2β   and   1 + cot2β = coesce2β ]

⇒ (1 – cos2β)(1 – sin2β) × cosec2β × sec2β

[ ∵   sin2α = sin2(90 – β) = cos2β    and  cos2α = cos2(90 – β) = sin2β ]

 ∴   sin2β . cos2β × 1 × 1 = 1 cos2β sin2β

[ ∵   (1 – cos2β) = sin2β   and   (1 – sin2β) = cos2β ]

Hence, option A is correct.

3
If α and β are complementary angles, then what is  cosα cosecβ – cosα sinβ  equal to?
» Explain it
C
α & β complementary angle.

α = 90 – β & β = 90 – α

cosα.cosecβ – cosα.sinβ

= cosα.cosec(90 – α) – cosα.sin(90 – α)

[ ∵ cosec(90 – α) = secα and sin(90 – α) = cosα ]

cosα.secα – cosα.cosα

 [ ∵ cosα.secα = cosα × 1 = 1 ] cosα

= 1 – cos2α

= sin2α = sinα.

Hence, option C is correct.

4
The value of cos 25° – sin 25° is
» Explain it
A
Since, value of cos Θ decreases, from 0° to 90° and at 45° it is equal to the value of sin Θ.

Similarly, value of sin Θ increases from 0° to 90° and at 45° it is equal to the value of cos Θ.

For 0° < Θ < 45°, cos Θ > sin Θ

So, value of cos 25° – sin 25° is always positive but less than 1.

Hence, option A is correct.

5
If sin (A + B) = 1, where 0 < B < 45°, then what is cos (A – B) equal to?
» Explain it
A

sin (A + B) = 1

⇒ A + B = sin–1 1

⇒ (A + B) = 90°

⇒ B = 90° – A

⇒ A = 90° – B

Now, cos (A – B) = cos A cos B + sin A + sin B

= cos (90 – B) cos B + sin (90 – B) sin B

= sin B cos B + cos B sin B

= 2 sin B cos B = sin 2B.

Hence, option A is correct.