Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Let the given expression be $E$.$$E=\frac{1+\sin (2\theta )+\cos (2\theta )}{1+\sin (2\theta )-\cos (2\theta )}$$ Using the double angle identities:$1+\cos (2\theta )=2{\cos }^2\theta$, $1-\cos (2\theta )=2{\sin }^2\theta$, and $\sin (2\theta )=2\sin \theta \cos \theta$. Substitute these into the expression:$$E=\frac{(1+\cos (2\theta ))+\sin (2\theta )}{(1-\cos (2\theta ))+\sin (2\theta )}=\frac{2{\cos }^2\theta +2\sin \theta \cos \theta }{2{\sin }^2\theta +2\sin \theta \cos \theta }$$ Factor out$2\cos \theta$ from the numerator and $2\sin \theta$ from the denominator:$$E=\frac{2\cos \theta (\cos \theta +\sin \theta )}{2\sin \theta (\sin \theta +\cos \theta )}$$ Assuming$\sin \theta +\cos \theta \ne 0$, we can cancel the common term:$$E=\frac{\cos \theta }{\sin \theta }=\cot \theta$$

Now we need to express$\cot \theta$ in terms of $P$ and $Q$. We know $\cot \theta =\frac{\cos \theta }{\sin \theta }$. Multiply numerator and denominator by$\cos \theta$ (or $\sin \theta$):$$\cot \theta =\frac{{\cos }^2\theta }{\sin \theta \cos \theta }$$ Using${\cos }^2\theta =\frac{1+\cos (2\theta )}{2}$ and $\sin \theta \cos \theta =\frac{\sin (2\theta )}{2}$:$$\cot \theta =\frac{(1+\cos (2\theta ))/2}{\sin (2\theta )/2}=\frac{1+\cos (2\theta )}{\sin (2\theta )}$$ Substitute$P=\cos (2\theta )$ and $Q=\sin (2\theta )$:$$E=\frac{1+P}{Q}$$