Question

$$\begin{gathered} Q.Provethat:se{c}^4\theta -se{c}^2\theta ={\tan }^4\theta +{\tan }^2\theta \\ Q.FindacuteanglesAandB,if \\ \sin (A+2B)=\frac{\sqrt{3}}{2}and\cos \left(A+4B\right)=0,A>B \end{gathered}$$

Answer 1

$$\begin{gathered} \left(1\right). \\ \mathrm{LHS}={\sec }^4\mathrm{\theta }-{\sec }^2\mathrm{\theta } \\ ={\sec }^2\mathrm{\theta }\left({\sec }^2\mathrm{\theta }-1\right) \\ =\left(1+{\tan }^2\mathrm{\theta }\right){\tan }^2\mathrm{\theta } \\ ={\tan }^2\mathrm{\theta }+{\tan }^4\mathrm{\theta } \\ =\mathrm{RHS} \\ \left(2\right). \\ \sin \left(\mathrm{A}+2\mathrm{B}\right)=\frac{\sqrt{3}}{2} \\ \Rightarrow \sin \left(\mathrm{A}+2\mathrm{B}\right)=\sin 60° \\ \Rightarrow \mathrm{A}+2\mathrm{B}=60°....\left(1\right) \\ \cos \left(\mathrm{A}+4\mathrm{B}\right)=0 \\ \Rightarrow \cos \left(\mathrm{A}+4\mathrm{B}\right)=\cos 90° \\ \Rightarrow \mathrm{A}+4\mathrm{B}=90°......\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 2\mathrm{B}=30°\Rightarrow \mathrm{B}=15° \\ \mathrm{Now},\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{A}+30°=60° \\ \Rightarrow \mathrm{A}=30° \end{gathered}$$