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Direction: Study the following questions carefully and choose the right answer:
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
      [ ∵ sec Θ = 
1
cos Θ
     &     tan Θ = 
sin Θ
cos 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 , ∠C =   π  and D divides BC internally in the ratio 1 : 3 then   sin ∠BAD  is equal to
3 4 sin ∠CAD

» Explain it
C
 
Given that ∠B =  π  and ∠C =  π
3 4

and  BD  =  1
CD 3

In ΔABD,

BD  =  AD
sin ∠BAD sin ∠ABD

⇒   BD  =  AD
sin ∠BAD sin π/3

⇒   BD  =  AD
sin ∠BAD
3 /2

⇒  sin ∠BAD = 
3
. BD        ...(i)
2 AD

In ΔACD,

CD  =  AD
sin ∠CAD sin ∠ACD

⇒   CD  =  AD
sin ∠CAD sin π/4

⇒   CD  =  AD
sin ∠CAD
1/ 2

⇒  sin ∠CAD =  1 . CD        ...(ii)
2 AD

Equation (i) ÷ (ii),

sin ∠BAD
sin ∠CAD
 = 
3
. BD
2 AD
1 . CD
2 AD

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  = sec2 Θ –  sin2 Θ (1 – 2 sin2 Θ)
2 cos4 Θ – cos2 Θ cos2 Θ (2 cos2 Θ – 1)

= sec2 Θ –  sin2 Θ cos 2Θ         [∵ cos 2Θ  =  1 – 2 sin2 Θ  =  2 cos2 Θ – 1]
cos2 Θ cos 2Θ

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