Math, asked by rajeshtinku890, 4 months ago

sinѲ+cosѲ/sinѲ-cosѲ + sinѲ-cosѲ/sinѲ+cosѲ= 2/2sin²Ѳ-1
help me please!!!!​

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Answered by stuti2526
4

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hope it helped you!!!!!!

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Answered by mysticd
3

 LHS = \frac{(sin \theta + cos \theta)}{(sin \theta - cos \theta )} +  \frac{(sin \theta - cos \theta)}{(sin \theta + cos \theta )} \\= \frac{(sin\theta + cos \theta )^{2} + (sin \theta - cos \theta)^{2}}{(sin \theta - cos \theta )(sin \theta + cos \theta)} \\= \frac{2(sin^{2} \theta + cos^{2}\theta)}{sin^{2} \theta - cos^{2} \theta } \\= \frac{ 2 \times 1 }{ sin^{2}\theta - (1 - sin^{2} \theta )}\\= \frac{2}{sin^{2} \theta - 1 + sin^{2} \theta }\\= \frac{2}{2sin^{2} \theta - 1 } \\= RHS

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