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Prove that --(cosec-sec)(cot-tan)=(cosec+sec)(sec*cosec-2) There is a theta after every trignometric term. I tried solving the lhs and rhs seperately but they were not coming out to be equal. Please help me answer this question. And also how does one go about proving such questions? Thanks in advance

Prove that --(cosec-sec)(cot-tan)=(cosec+sec)(sec*cosec-2)
There is a theta after every trignometric term. I tried solving the lhs and rhs seperately but they were not coming out to be equal.
Please help me answer this question. 
And also how does one go about proving such questions? 
Thanks in advance

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

(cosec\theta -sec\theta )(cot\theta -tan\theta ) = (cosec\theta +sec\theta )(cosec\theta sec\theta -2)
LHS = (cosec\theta -sec\theta )(cot\theta -tan\theta )
LHS = (\frac{1}{sin\theta} -\frac{1}{cos\theta} )(\frac{cos\theta}{sin\theta} -\frac{sin\theta}{cos\theta})
LHS = (\frac{cos\theta-sin\theta}{sin\theta.cos\theta})(\frac{cos^{2}\theta-sin^{2}\theta}{sin\theta.cos\theta})
LHS = (\frac{cos\theta-sin\theta}{sin\theta.cos\theta})(\frac{(cos\theta-sin\theta)(cos\theta+sin\theta)}{sin\theta.cos\theta})
LHS = (\frac{(cos\theta-sin\theta)^{2}(cos\theta+sin\theta)}{sin^{2}\theta.cos^{2}\theta})
LHS = (\frac{(cos^{2}\theta+sin^{2}\theta-2cos\theta.sin\theta)(cos\theta+sin\theta)}{sin^{2}\theta.cos^{2}\theta})
LHS = \frac{(cos\theta+sin\theta)(1-2sin\theta.cos\theta)}{sin^{2}\theta.cos^{2}\theta}
RHS = (cosec\theta +sec\theta )(cosec\theta sec\theta -2)
RHS = (\frac{1}{sin\theta} +\frac{1}{cos\theta} )(\frac{1}{sin\theta}.\frac{1}{cos\theta} -2)
RHS = (\frac{sin\theta+cos\theta}{sin\theta.cos\theta})(\frac{1-2sin\theta.cos\theta}{sin\theta.cos\theta})
RHS = \frac{(cos\theta+sin\theta)(1-2sin\theta.cos\theta)}{sin^{2}\theta.cos^{2}\theta}
LHS = RHS
Hence proved.

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