Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A

Asked by Abhisek | 1 year ago |  133

##### Solution :-

L.H.S. = (sin A + cosec A)+ (cos A + sec A)2

It is of the form (a+b)2, expand it

(a+b)2 =a2 + b2 +2ab

= (sin2A + cosec2A + 2 sin A cosec A)

+ (cos2A + sec2A + 2 cos A sec A)

= (sin2A + cos2A) + 2 sin A($$\dfrac{1}{sin A}$$

+ 2 cos A ($$\dfrac{1}{cos A}$$) + 1 + tan2A + 1 + cot2A

= 1 + 2 + 2 + 2 + tan2A + cot2A

= 7+tan2A+cot2A = R.H.S.

Therefore, (sin A + cosec A)+ (cos A + sec A)2

= 7+tan2A+cot2A

Hence proved.

Answered by Pragya Singh | 1 year ago

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