Trigonometry> Prove that --(cosec-sec)(cot-tan)=(cosec+...

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

Jai Mahajan , 11 Years ago

Grade 12

1 Answers

Jitender Singh

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.

Last Activity: 11 Years ago