Question

simplify (sin theta+cosec theta )^2+(cos theta+sec theta)^2-(tan theta+cot theta)^2

Answer 1

$\textbf{Given:}$

$\displaystyle(sin\;\theta+cosec\;\theta)^2+(cos \;\theta+sec\;\theta)^2-(tan\;\theta+cot \;\theta)^2$

$\text{Using the following algebraic identities}$

$\implies\boxed{\begin{minipage}{4cm}\mathsf{ \bf(a+b)^2=a^2+b^2+2ab\\\\(a-b)^2=a^2+b^2-2ab }\end{minipage}}$

$=\displaystyle(sin^2\theta+cosec^2\theta+2\;sin\;\theta\;cosec\;\theta)+(cos^2\theta+sec^2\theta+2\;cos\theta\;sec\theta)-(tan^2\theta+cot^2\theta+2\;tan\;\theta\;cot \;\theta)$

$=\displaystyle\;sin^2\theta+cosec^2\theta+2\;sin\;\theta(\frac{1}{sin\;\theta})+(cos^2\theta+sec^2\theta+2\;cos\theta(\frac{1}{cos\;\theta})-(tan^2\theta+cot^2\theta+2\;tan\;\theta(\frac{1}{cot\;\theta}))$

$=\displaystyle\;sin^2\theta+cosec^2\theta+2(1)+cos^2\theta+sec^2\theta+2(1)-(tan^2\theta+cot^2\theta+2(1))$

$=\displaystyle\;sin^2\theta+cosec^2\theta+2+cos^2\theta+sec^2\theta+2-tan^2\theta-cot^2\theta-2$

$=\displaystyle\;sin^2\theta+cosec^2\theta+2+cos^2\theta+sec^2\theta-tan^2\theta-cot^2\theta$

$=\displaystyle\;(sin^2\theta+cos^2\theta)+(sec^2\theta-tan^2\theta)+(cosec^2\theta-cot^2\theta)+2$

$=\displaystyle\;1+1+1+2$

$=\displaystyle\;5$

$\therefore\boxed{\bf(sin\;\theta+cosec\;\theta)^2+(cos \;\theta+sec\;\theta)^2-(tan\;\theta+cot \;\theta)^2=5}$