Q1. Divide by $\cos \theta$ to get $\tan \theta + 1 = \sqrt{2} \rightarrow \tan \theta = \sqrt{2}-1. Show LHS simplifies to same relation.
Q2. Substitute values: $\frac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} = \frac{5(1/4) + 4(4/3) - 1}{1/4 + 3/4} = \frac{5/4 + 16/3 - 1}{1} = \frac{15 + 64 - 12}{12} = \frac{67}{12}$.
Q3. Convert into sine and cosine: LHS simplifies to $\frac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}$ and expand numerator using $a^3-b^3$.
Q4. $p = \sec\theta + \tan\theta \rightarrow 1/p = \sec\theta - \tan\theta. Solve for $\sec\theta$ and $\tan\theta, then $\sin\theta = \tan\theta/\sec\theta$.
Q5. Multiply numerator and denominator inside root by $(1+\sin A). It becomes $\sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} = \frac{1+\sin A}{\cos A} = \sec A + \tan A$.
