Question

Sin square theta tan theta + cos square theta cot theta + 2 Sin Theta multiplied cos theta equal 10 theta + cot theta = sec theta x cos theta

Answer 1

Answer:

To prove: sin^2\,\theta\:tan\,\theta+cos^2\,\theta\:cot\,\theta+2sin\,\theta\:cos\,\theta=tan\,\theta+cot\,\theta

Consider,

LHS

=sin^2\,\theta\:tan\,\theta+cos^2\,\theta\:cot\,\theta+2sin\,\theta\:cos\,\theta

=sin^2\,\theta\:tan\,\theta+cos^2\,\theta\:cot\,\theta+sin\,\theta\:cos\,\theta+sin\,\theta\:cos\,\theta

=sin^2\,\theta\:tan\,\theta+sin\,\theta\:cos\,\theta+cos^2\,\theta\:cot\,\theta+sin\,\theta\:cos\,\theta

=sin\,\theta(sin\,\theta\:tan\,\theta+cos\,\theta)+cos\,\theta(cos\,\theta\:cot\,\theta+sin\,\theta)

=sin\,\theta(sin\,\theta\times\frac{sin\,\theta}{cos\,\theta}+cos\,\theta)+cos\,\theta(cos\,\theta\times\frac{cos\,\theta}{sin\,\theta}+sin\,\theta)

=sin\,\theta(\frac{sin^2\,\theta+cos^2\,\theta}{cos\,\theta})+cos\,\theta(\frac{cos^\,\theta+sin^2\,\theta}{sin\,\theta})

=sin\,\theta(\frac{1}{cos\,\theta})+cos\,\theta(\frac{1}{sin\,\theta})

=\frac{sin\,\theta}{cos\,\theta}+\frac{cos\,\theta}{sin\,\theta}

=tan\,\theta+cot\,\theta

= RHS

Hence Proved