Question

prove that cos theta by 1 + tan theta + sin square theta by sin theta minus cos theta is equal to sin theta + cos theta​

Answer 1

Answer:

Step-by-step explanation:

\frac{sin\theta+cos\theta}{sin\theta-cos\theta}+\frac{sin\theta-cos\theta}{sin\theta+cos\theta}

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

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

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

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

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

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

=\frac{2}{sin^{2}\theta-(1-sin^{2}\theta)}

=\frac{2}{sin^{2}\theta-1+sin^{2}\theta}

=\frac{2}{2sin^{2}\theta-1}