Question

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

Answer 1

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

Answer 2

Answer:

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