R K Bansal Solutions for Exercise 1: EXERCISE

If $\mathrm{sinA}+\mathrm{cosA}=\mathrm{p}$ and $\mathrm{secA}+\mathrm{cosecA}=\mathrm{q}$, then prove that: $\mathrm{q}\left({\mathrm{p}}^2-1\right)=2\mathrm{p}$.

Find $\mathrm{A}$, if $0°\le \mathrm{A}\le 90°$ and ${\cos }^2\mathrm{A}-\mathrm{cosA}=0$

Find $\mathrm{A}$, if $0°\le \mathrm{A}\le 90°$ and $2{\cos }^2\mathrm{A}+\mathrm{cosA}-1=0$

If $0°<\mathrm{A}<90°$; find $\mathrm{A}$ if $\frac{\mathrm{cosA}}{1-\mathrm{sinA}}+\frac{\mathrm{cosA}}{1+\mathrm{sinA}}=4$

If $0°<\mathrm{A}<90°$; find $\mathrm{A}$ if $\frac{\mathrm{sinA}}{\mathrm{secA}-1}+\frac{\mathrm{sinA}}{\mathrm{secA}+1}=2$

Prove that $(cosec\mathrm{A}-\mathrm{sinA})(\mathrm{secA}-\mathrm{cosA}){\sec }^2\mathrm{A}=\mathrm{tanA}$

Prove that :

$(\mathrm{sin\theta }+\mathrm{cos\theta })(\mathrm{tan\theta }+\mathrm{cot\theta })=\mathrm{sec\theta }+cosec\mathrm{\theta }$

Prove that :

${\sin }^{2}28°+{\sin }^{2}62°+{\tan }^{2}38°-{\cot }^{2}52°+\frac{1}{4}{\sec }^{2}30°$