If tan theta - cot theta = a and sin theta + | vedclass

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Question

(d) Given that $	an 	heta - \cot 	heta = a$&hellip;..$(i)$

and $\sin 	heta + \cos 	heta = b$&hellip;..$(ii)$

Now ${({b^2} - 1)^2}({a^2} + 4

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) Given that $	an 	heta - \cot 	heta = a$&hellip;..$(i)$

and $\sin 	heta + \cos 	heta = b$&hellip;..$(ii)$

Now ${({b^2} - 1)^2}({a^2} + 4)$

$ = {\left\{ {{{(\sin 	heta + \cos 	heta )}^2} - 1} ight\}^2}\left\{ {{{(	an 	heta - \cot 	heta )}^2} + 4} ight\}$

$ = {[1 + \sin 2	heta - 1]^2}[{	an ^2}	heta + {\cot ^2}	heta - 2 + 4]$

$ = {\sin ^2}2	heta ({m{cose}}{{m{c}}^2}	heta + {\sec ^2}	heta )$

$ = 4{\sin ^2}	heta {\cos ^2}	heta \left[ {\frac{1}{{{{\sin }^2}	heta }} + \frac{1}{{{{\cos }^2}	heta }}} ight] = 4$. 

Trick : Obviously the value of expression ${({b^2} - 1)^2}({a^2} + 4)$ is independent of $	heta $,

therefore put any suitable value of $	heta $.

Let $	heta = 45^\circ $, we get $a = 0,\;b = \sqrt 2 $

so that ${[{(\sqrt 2 )^2} - 1]^2}$ $({0^2} + 4) = 4.$

If $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b,$ then ${({b^2} - 1)^2}({a^2} + 4)$ is equal to

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