Question

prove that :. (sin∅)/ ( sec∅+ tan∅-1) +. ( cos∅) /( cosec∅+ cot∅-1) = 1​

Answer 1

Answer:

LHS:

$\\\\\\\tt \dfrac{\sin \: \theta}{ \sec\theta + \tan\theta - 1 } + \dfrac{ \cos\theta}{\cosec\theta + \cot\theta - 1}$

$\tt\dfrac{\sin\theta}{\sec\theta + \tan\theta - ( {\sec}^{2}\theta - {\tan}^{2}\theta) } + \dfrac{\cos\theta}{\cosec\theta + \cot\theta - ( {\cosec}^{2}\theta - {\cot}^{2}\theta) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\\ \tt \dfrac{\sin\theta}{\sec\theta + \tan\theta -( ( {\sec}\theta + \tan\theta)(\sec \theta - \tan\theta)) } + \dfrac{\cos\theta}{\cosec\theta + \cot\theta -( (\cosec\theta + \cot\theta) (\cosec\theta - \cot\theta))}$

$\tt\dfrac{\sin\theta}{(\sec\theta + \tan\theta )(1 - (\sec\theta - \tan\theta)) } + \dfrac{\cos\theta}{(\cosec\theta + \cot\theta)(1 -(\cosec\theta - \cot\theta)) } \\\\ \tt\dfrac{\sin\theta}{(\sec\theta + \tan\theta )(1 - \sec\theta + \tan\theta) } + \dfrac{\cos\theta}{(\cosec\theta + \cot\theta)(1 - \cosec\theta + \cot\theta) }$

Now converting every term in sin and cos

$\\\\\tt\dfrac{\sin\theta}{( \dfrac{1}{\cos \theta} + \dfrac{\sin\theta}{\cos \theta })(1 -\dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta}) } + \dfrac{\cos\theta}{( \dfrac{1}{\sin\theta} + \dfrac{\cos\theta}{\sin\theta})(1 - \dfrac{1}{\sin\theta } + \dfrac{\cos\theta}{\sin\theta}) }\\\\\tt\dfrac{\sin\theta}{( \dfrac{1 + \sin\theta}{\cos\theta} )( \dfrac{\cos\theta - 1 + \sin\theta}{\cos\theta} ) } + \dfrac{\cos\theta}{( \dfrac{1 + \cos\theta}{\sin\theta} )( \dfrac{\sin\theta - 1 + \cos\theta}{\sin\theta} ) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt\dfrac{\sin\theta {\cos}^{2}\theta }{(1 + \sin\theta)(\cos\theta - 1 + \sin\theta) } + \dfrac{\cos\theta {\sin}^{2}\theta }{(1 + \cos\theta)(\sin\theta - 1 + \cos\theta) }\\ \\ \tt\dfrac{\sin\theta(1 - {\sin}^{2}\theta) }{(1 + \sin\theta)(\sin\theta + \cos\theta - 1) } + \dfrac{\cos\theta(1 - {\cos}^{2} \theta}{(1 + \cos\theta )(\sin\theta + \cos\theta - 1) }$

$\tt\dfrac{\sin\theta(1 + \sin\theta)(1 - \sin\theta) }{(1 + \sin\theta)(\sin\theta + \cos\theta - 1) } + \dfrac{\cos\theta(1 + \cos \theta)(1 - \cos\theta)}{(1 + \cos\theta )(\sin\theta + \cos\theta - 1) }$

$\tt\dfrac{(\sin\theta - {\sin}^{2}\theta) }{(\sin\theta + \cos\theta - 1)} + \dfrac{(\cos\theta- {\cos}^{2} \theta}{(\sin\theta + \cos\theta - 1) )}$

$\tt\dfrac{\sin\theta+\cos\theta-{\sin}^2\theta-{\cos}^2\theta}{\sin\theta+\cos\theta-1}$

$\tt\dfrac{\sin\theta+\cos\theta-(\sin^2\theta+\cos^2\theta)}{\sin\theta+\cos\theta-1}$

Substituting the value of $\sf\sin^2\theta+\cos^2\theta$ as 1.

$\\\\\tt\dfrac{\sin\theta+\cos\theta-1}{\sin\theta+\cos\theta-1}$

On cancelling, we get:

$\\\\\tt1$

RHS:

$\\\\\\1$

LHS = RHS

Hence Proved.