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Question Number 14991 by tawa tawa last updated on 06/Jun/17

Answered by ajfour last updated on 06/Jun/17

$$=\frac{\tan \theta }{(1-\frac{1}{\tan \theta })}+\frac{1}{\tan \theta (1-\tan \theta )}$$$$=\frac{\tan {}^2\theta }{\tan \theta -1}-\frac{1}{\tan \theta (\tan \theta -1)}$$$$=\frac{\tan {}^3\theta -1}{\tan \theta (\tan \theta -1)}$$$$=\frac{\tan {}^2\theta +1+\tan \theta }{\tan \theta }(if\tan \theta \ne 1)$$$$=\frac{\sec {}^2\theta +\tan \theta }{\tan \theta }=\frac{\sec {}^2\theta }{\tan \theta }+1$$$$=\frac{1}{\left(\cos {}^2\theta \right)\left(\frac{\sin \theta }{\cos \theta }\right)}+1$$$$=\frac{1}{\sin \theta \cos \theta }+1$$$$=\mathrm{cosec}\theta \sec \theta +1.$$

Commented by tawa tawa last updated on 06/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by RasheedSoomro last updated on 06/Jun/17

$$\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \csc \theta$$$$\mathrm{LHS}:\frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\sin \theta }{\cos \theta }}$$$$=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin \theta -\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta -\sin \theta }{\cos \theta }}$$$$=\frac{\sin \theta }{\cos \theta }\times \frac{\sin \theta }{\sin \theta -\cos \theta }+\frac{\cos \theta }{\sin \theta }\times \frac{\cos \theta }{\cos \theta -\sin \theta }$$$$=\frac{{\sin }^2\theta }{\cos \theta (\sin \theta -\cos \theta )}+\frac{{\cos }^2\theta }{\sin \theta (\cos \theta -\sin \theta )}$$$$=\frac{1}{\sin \theta -\cos \theta }(\frac{{\sin }^2\theta }{\cos \theta }-\frac{{\cos }^2\theta }{\sin \theta })$$$$=\frac{1}{\sin \theta -\cos \theta }\left(\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }\right)$$$$=\frac{1}{\sin \theta -\cos \theta }\times \frac{(\sin \theta -\cos \theta )({\sin }^2\theta +\sin \theta \cos \theta +{\cos }^2\theta )}{\sin \theta \cos \theta }$$$$=\frac{{\sin }^2\theta +\sin \theta \cos \theta +{\cos }^2\theta }{\sin \theta \cos \theta }$$$$=\frac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }$$$$=\frac{1}{\sin \theta \cos \theta }+\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }$$$$=1+\frac{1}{\cos \theta }\times \frac{1}{\sin \theta }$$$$=1+\sec \theta \csc \theta =\mathrm{RHS}$$$$$$

Commented by tawa tawa last updated on 06/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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