Question

Prove the following trigonometric identities.

${\displaystyle {[\tan \theta +\frac{1}{\cos \theta }]}^2+{[\tan \theta -\frac{1}{\cos \theta }]}^2=2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)}$

Answer 1

LHS = (tan θ + sec θ)²  + (tan θ  - sec θ)²

${\displaystyle \text{LHS}={\tan }^2\theta +{\sec }^2\theta +\cancel{2\tan \theta .\sec \theta }+ {\tan }^2\theta +{\sec }^2\theta -\cancel{2\tan \theta .\sec \theta }...\{\begin{array}{l}{a}^2+b^2=a^2+2ab+b^2\\ a^2-b^2=a^2-2ab+b^2\end{array}}$

LHS = ${\displaystyle 2{\tan }^2\theta +2{\sec }^2\theta }$

LHS = ${\displaystyle 2[{\tan }^2\theta +{\sec }^2\theta ]}$

LHS = ${\displaystyle 2[\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{1}{{\cos }^2\theta }]}$  

LHS = ${\displaystyle 2\left(\frac{{\sin }^2\theta +1}{{\cos }^2\theta }\right)}$

LHS = ${\displaystyle 2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)...\{\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ \therefore {\cos }^2\theta =1-{\sin }^2\theta \end{array}}$

RHS = ${\displaystyle 2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)}$

LHS = RHS

Answer 2

LHS = ${\displaystyle {[\tan \theta +\frac{1}{\cos \theta }]}^2+{[\tan \theta -\frac{1}{\cos \theta }]}^2 ...\{{(a+b)}^2+{(a-b)}^2=2(a^2+b^2)\}}$

LHS = ${\displaystyle 2[{\tan }^2\theta +\frac{1}{{\cos }^2\theta }]}$

LHS = ${\displaystyle 2[\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{1}{{\cos }^2\theta }]}$

LHS = ${\displaystyle 2\left(\frac{{\sin }^2\theta +1}{{\cos }^2\theta }\right)}$

LHS = ${\displaystyle 2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)...\{\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ \therefore {\cos }^2\theta =1-{\sin }^2\theta \end{array}}$

RHS = ${\displaystyle 2\left(\frac{1+{\sin }^2\theta }{1-{\sin }^2\theta }\right)}$

LHS = RHS