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Question Number 172124 by Mikenice last updated on 23/Jun/22

$$iftan\theta +sec\theta =x,showthat$$$$sin\theta =\frac{x^2-1}{x^2+1}$$

Commented by infinityaction last updated on 23/Jun/22

$$\frac{{(1+\sin \theta )}^2}{{\cos }^2\theta }=x^2$$$$\frac{1+{\sin }^2\theta +2\sin \theta }{{\cos }^2\theta }=x^2$$$$\frac{1+{\sin }^2\theta +2\sin \theta -{\cos }^2\theta }{1+{\sin }^2\theta +2\sin \theta +{\cos }^2\theta }=\frac{x^2-1}{x^2+1}$$$$\frac{\cancel{2sin}\theta \cancel{(1+\sin \theta )}}{\cancel{2}\cancel{(1+\sin \theta )}}=\frac{x^2-1}{x^2+1}$$$$\sin \theta =\frac{x^2-1}{x^2+1}$$

Commented by BaliramSingh last updated on 24/Jun/22

$$NiceSolution$$

Answered by BaliramSingh last updated on 23/Jun/22

$$$$$$sec\theta +tan\theta =x..........\left[1\right]$$$$\frac{(sec\theta +tan\theta )(sec\theta -tan\theta )}{(sec\theta -tan\theta )}=x$$$$\frac{{sec}^2\theta -{tan}^2\theta }{(sec\theta -tan\theta )}=x$$$$\frac{1}{(sec\theta -tan\theta )}=x$$$$sec\theta -tan\theta =\frac{1}{x}...........\left[2\right]$$$$by\left[1\right]+\left[2\right]$$$$2sec\theta =x+\frac{1}{x}............\left[3\right]$$$$by\left[1\right]-\left[2\right]$$$$2tan\theta =x-\frac{1}{x}.............\left[4\right]$$$$by\left[4\right]\mathbf{÷}\left[3\right]$$$$\frac{2tan\theta }{2sec\theta }=\frac{x-\frac{1}{x}}{x+\frac{1}{x}}$$$$\frac{\frac{sin\theta }{cos\theta }}{\frac{1}{cos\theta }}=\frac{\frac{x^2-1}{x}}{\frac{x^2+1}{x}}$$$$sin\theta =\frac{x^2-1}{x^2+1}$$$$$$

Commented by peter frank last updated on 23/Jun/22

$$\mathrm{thank}\mathrm{you}$$

Answered by Plato last updated on 23/Jun/22

$$soln$$$$\sin \theta =\frac{x^2-1}{x^2+1}$$$$\frac{x^2-1}{x^2+1}$$$$\frac{{(\tan \theta +\sec \theta )}^2-1}{{(\tan \theta +\sec \theta )}^2+1}$$$$\frac{\tan {}^2\theta +\sec {}^2\theta +2\tan \theta \sec \theta -1}{\tan {}^2\theta +\sec {}^2\theta +2\tan \theta \sec \theta +1}$$$$but1+{\tan }^2\theta =\sec {}^2\theta$$$$\frac{2\tan \theta \sec \theta +{2\tan }^2\theta }{2\tan \theta \sec \theta +2\tan {}^2\theta +2}$$$$factoring2$$$$\frac{\tan \theta \sec \theta +{\tan }^2\theta }{\tan \theta \sec \theta +{\sec }^2\theta }$$$$divideby\sec \theta$$$$\frac{\tan \theta +{\tan }^2\theta \cos \theta }{\tan \theta +\frac{1}{\cos \theta }}$$$$\frac{\frac{\sin \theta }{\cos \theta }+\frac{{\sin }^2\theta }{\cos \theta }}{\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }}$$$$\frac{\frac{\sin \theta (1+\sin \theta )}{\cos \theta }}{\frac{1+\sin \theta }{\cos \theta }}$$$$\frac{\sin \theta (1+\sin \theta )}{(1+\sin \theta )}$$$$=\sin \theta$$

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