Question

Find $\left(\cos A+\sec A\right){}^2+{\left(\sin A+\text{cosec}A\right)}^2-{\tan }^{2}A-{\cot }^{2}A.$

Answer 1

$(\cos A+\sec A{)}^2+(\sin A+\csc A{)}^2-{\tan }^{2}A-{\cot }^{2}A$

$\Rightarrow \sec A=\frac{1}{\cos A}$

$\Rightarrow \csc A=\frac{1}{\sin A}$

$\Rightarrow \tan A=\frac{\sin A}{\cos A}$

$\Rightarrow \cot A=\frac{\cos A}{\sin A}$

$\Rightarrow$ put value in expand:

$\Rightarrow {\cos }^{2}A+\frac{1}{{\cos }^{2}A}+2+{\sin }^{2}A+\frac{1}{{\sin }^{2}A}+2-\frac{{\sin }^{2}A}{{\cos }^{2}A}-\frac{co{s}^{2}A}{{\sin }^{2}A}$

$\Rightarrow 4+{\cos }^{2}A+{\sin }^{2}A+\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A}-\frac{{\sin }^{2}A}{{\cos }^{2}A}-\frac{{\cos }^{2}A}{si{n}^{2}A}$

$\Rightarrow \therefore {\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow 4+1+\frac{{\cos }^{2}A+{\sin }^{2}A}{{\sin }^{2}A{\cos }^{2}A}-\frac{{\sin }^{4}A+{\cos }^{4}A}{{\sin }^{2}A{\cos }^{2}A}$

$\Rightarrow 5+\frac{1}{{\sin }^{2}A{\cos }^{2}A}-\frac{({\sin }^{2}A{)}^2+({\cos }^{2}A{)}^2}{{\sin }^{2}A{\cos }^{2}A}$

$\Rightarrow \because a^2+b^2=(a+b{)}^2-2ab$

$\Rightarrow 5+\frac{1}{{\sin }^{2}A{\cos }^{2}A}-\frac{({\sin }^{2}A+{\cos }^{2}A{)}^2-2{\sin }^{2}A{\cos }^{2}A}{\sin 2A{\cos }^{2}A}$

$\Rightarrow 5+\frac{1}{{\sin }^{2}A{\cos }^{2}A}-\frac{1-2{\sin }^{2}A{\cos }^{2}A}{{\sin }^{2}A{\cos }^{2}A}$

$\Rightarrow 5+\frac{1-1+2{\sin }^{2}A{\cos }^{2}A}{{\sin }^{2}A{\cos }^{2}A}$

$\Rightarrow 5+2=7$