Question

If sec theta + tan theta =x prove that sin tjeta =x²-1/x²+1

Answer 1

Hope this answer may help

Answer 2

Correct Question:-

$\sf If \blue{ sec\theta+tan\theta=x} \: then \: prove \: that \: \red{ sin \theta= \dfrac{ {x}^{2} - 1 }{ {x}^{2} + 1} }$

Identities used

$\sf\pink {\bigstar{ {sec}^{2} \theta - {tan}^{2} \theta = 1 }}$

$\green{ \sf \bigstar \: tan \theta = \dfrac{sin \theta}{cos \theta} }$

$\orange{ \sf \bigstar \: \dfrac{1}{sec \theta} = cos \theta }$

Solution:-

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We have R•H•S

$\implies \sf \dfrac{ {x}^{2} - 1 }{ {x}^{2} + 1}$

Now putting the value of x in above

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$\sf \implies \dfrac{{( sec \theta +tan \theta ) }^{2} - 1}{{( sec \theta +tan \theta ) }^{2} + 1}$

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$\sf \implies \dfrac{ {sec}^{2} \theta + tan^{2} \theta + 2sec \theta.tan \theta - 1}{ {sec}^{2} \theta + tan^{2} \theta + 2sec \theta.tan \theta + 1}$

Now,by using below trignometric identity ,we get

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$\sf\pink {\bigstar{ {sec}^{2} \theta - {tan}^{2} \theta = 1 }}$

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$\sf \implies \dfrac{ {sec}^{2} \theta + tan^{2} \theta + 2sec \theta.tan \theta -{ ({sec}^{2} \theta - {tan}^{2} \theta ) }}{ {sec}^{2} \theta + tan^{2} \theta + 2sec \theta.tan \theta +{ ({sec}^{2} \theta - {tan}^{2} \theta ) } }$

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$\sf \implies \dfrac{ \cancel{{sec}^{2} } \theta + tan^{2} \theta + 2sec \theta.tan \theta -{ \cancel{ {sec}^{2} \theta } + {tan}^{2} \theta }}{ {sec}^{2} \theta + \cancel{ tan^{2} \theta }+ 2sec \theta.tan \theta +{ {sec}^{2} \theta - \cancel{{tan}^{2} \theta} } }$

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$\sf \implies \dfrac{ tan^{2} \theta + 2sec \theta.tan \theta + {tan}^{2} \theta }{ {sec}^{2} \theta + 2sec \theta.tan \theta + {sec}^{2} \theta }$

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$\sf \implies \dfrac{ 2 tan^{2} \theta + 2sec \theta.tan \theta}{ 2{sec}^{2} \theta + 2sec \theta.tan \theta }$

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$\sf \implies \dfrac{ \cancel{2} ( tan^{2} \theta + sec \theta.tan \theta)}{ \cancel{ 2}({sec}^{2} \theta + sec \theta.tan \theta )}$

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$\sf \implies \dfrac{ {tan \theta} \cancel{( tan \theta + sec \theta)}}{sec \theta \cancel{({sec} \theta +tan \theta )} }$

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$\sf \implies \: tan \theta \times \dfrac{1}{sec \theta}$

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$\implies \sf \dfrac{sin \theta}{ \cancel{cos \theta}} \times \cancel{cos \theta}$

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$\sf \implies \red{ \sin \theta}$

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$\sf \implies \blue{L.H.S}$

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Hence,proved