Question

cos 2 theta/(1-tan theta)+sin 3 theta/(sin theta - cos theta)=1+sin theta.cos theta

Answer 1

hope this will help.

Answer 2

Solution:

LHS = $\frac{cos^{2}\theta}{(1-tan\theta)}+\frac{sin^{3}\theta}{(sin\theta-cos\theta}$

= $\frac{cos^{2}\theta}{1-\frac{sin\theta}{cos\theta}}+\frac{sin^{3}\theta}{(sin\theta-cos\theta)}$

= $\frac{cos^{2}\theta}{\frac{(cos\theta-sin\theta)}{cos\theta}}+\frac{sin^{3}\theta}{(sin\theta-cos\theta)}$

= $\frac{cos^{3}\theta}{(cos\theta-sin\theta)}-\frac{sin^{3}\theta}{(cos\theta-sin\theta)}$

= $\frac{cos^{3}\theta-sin^{3}\theta}{(cos\theta-sin\theta)}$

/* we know the algebraic identity:

a³-b³ = (a-b)(a²+ab+b²) */

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

After cancellation, we get

= $cos^{2}\theta+sin^{2}\theta}+sin\theta cos\theta$

/* We know the Trigonometric identity:

sin²A + cos²A = 1 */

= $1+sin\theta cos\theta$

= RHS

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