Question

show that 1 + cos theta minus sin square theta by sin theta 1 + cos theta equal to cot theta

Answer 1

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Answer 2

Solution:

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

= $\frac{[(1-sin^{2}\theta)+cos\theta]}{sin\theta(1+cos\theta)}$

/* We know the Trigonometric identity:

$\boxed {1-sin^{2}\theta = cos^{2}\theta}$ */

= $\frac{(cos^{2}\theta+cos\theta)}{sin\theta(1+cos\theta)}$

= $\frac{[cos\theta(1+cos\theta)]}{sin\theta(1+cos\theta)}$

After cancellation, we get

= $\frac{cos\theta}{sin\theta}$

= $cot\theta$

= $RHS$

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