Question

Solve it please
answer only ​

Answer 1

PROVE THAT -:

${ \sf{(cot \theta - cosec \theta) }^{2} = \frac{1 - cos \theta}{1 + cos \theta} }$

SOLUTION

L.H.S

${ \sf{(cot \theta - cosec \theta) }^{2} } \\ \\ \sf{ \implies \: \ { (\frac{1}{sin \theta} - \frac{cos \theta}{sin \theta}) }^{2} } \\ \\ \sf{ \implies \: ({ \frac{1 - cos \theta}{sin \theta} })^{2} } \\ \\ \sf{ \implies \: \frac{ {(1 - cos \theta})^{2} }{ {(sin \theta})^{2} } } \\ \\ \sf{ \implies\frac{ {(1 - cos \theta)}^{2} }{(1 - { {cos}^{2} \theta} )} } \\ \\ \sf{ \implies \frac{(1 - cos \theta)(1 - cos \theta)}{(1 + cos \theta)(1 - cos \theta)} } \\ \\ \sf{ \implies \: \frac{(1 - cos \theta)}{(1 + cos \theta) } =R.H.S }$

Hence proved R.H.S= L.H.S.

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Important points !!

✏️tan∅= p/b

✏️cos∅= b/h

✏️sin∅= p/h

✏️sin∅= 1/ cosec∅

✏️ cos∅= 1/sec∅

✏️sin²∅+ cos²∅= 1

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Answer 2

your answers

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