If \frac{ \sec^2 70 −\cot^2 20^\circ }{2 (cosec 5^2 59^\circ −\tan^2 31^\circ ) }=m/2 , then m is equal to

SSC CHSL 20 Dec 2015 Shift 2 Paper

Question : 136 of 200

Marks: +1, -0

{ sec^2 70 −cot^2 20° }/{2 (cosec 5^2 59° −tan^2 31° ) }=m/2

, then m is equal to

Solution:

Expression :

{ sec^2 70°−cot^2 20° }/{2 (cosec ^2 59° −tan^2 31° ) }=m/2

Using,

cot (90°=θ)=tanθ

and vice-versa
=

{ sec^2 70° −cot^2 (90°-70°) }/{2[ cosec ^2 59° −tan^2 (90°-59° ) ]}=m/2

{ sec^2 70 °− tan^2 70° }/{2 (cosec ^2 59° − cot^2 59° ) }=m/2

Using,

(sec^2θ-tan^2θ=1)

and

(cosec^2θ-cot^2θ=1)

1/{2×1} =m/ 2

⇒ m = 1

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