Math, asked by ranjanrshah55, 2 months ago

if tan^2 theta =8/7 then the value of (1+sin)(1-sin)/ (1-cos)(1+cos) ​

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Answered by manu5024
2

Answer:

This is your question's solution.

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Answered by Anonymous
4

Answer:

\sf \dfrac{(1 +  \sin\theta)(1 - \sin\theta) }{(1 +  \cos\theta )(1 -  \cos\theta)}  =  \dfrac{7}{8}

Step-by-step explanation:

\sf \dfrac{(1 +  \sin\theta)(1 - \sin\theta) }{(1 +  \cos\theta )(1 -  \cos\theta)}   =  \dfrac{1 -  { \sin}^{2}\theta }{1 -  { \cos }^{2}\theta}  \\  \\ \sf \dfrac{(1 +  \sin\theta)(1 - \sin\theta) }{(1 +  \cos\theta )(1 -  \cos\theta)}   =  \frac{ { \cos }^{2}\theta }{ { \sin}^{2}\theta}  \\  \\ \sf \dfrac{(1 +  \sin\theta)(1 - \sin\theta) }{(1 +  \cos\theta )(1 -  \cos\theta)}   =  { \cot }^{2} \theta =  \frac{1}{ { \tan}^{2} \theta}

\sf \dfrac{(1 +  \sin\theta)(1 - \sin\theta) }{(1 +  \cos\theta )(1 -  \cos\theta)}   =  \dfrac{1}{ \dfrac{8}{7} }  \\  \\ \sf \dfrac{(1 +  \sin\theta)(1 - \sin\theta) }{(1 +  \cos\theta )(1 -  \cos\theta)}   =  \dfrac{7}{8}

Hope it helps

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