Question

tan/1-cot + cot/1-tan = 1+seccosec

Answer 1

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Replace tan A by sin A/cos A and cot A by cos A/sin A. We get

[sin A / cos A]/[1 – cos A/sin A] + [cos A/sin A]/[1 – sin A/cos A]

Or sin A.sin A/[cos A(sin A – cosA)] + cos A.cos A/[sin A(cos A-sinA)].

LCM of denominator is sin A.cos A (sin A – cos A)

On simplifying we get

(sin^3 A – cos^3 A)/ [sin A.cos A (sin A – cos A)]

= (sin A – cos A)( sin^2 A + cos^2 A + sin A.cos A] / [sin A.cos A (sin A – cos A)]

= (sin A – cos A)( 1 + sin A.cos A] / [sin A.cos A (sin A – cos A)]

=( 1 + sin A.cos A] / sin A.cos A

= 1 + sec A.cosec A

Proved

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