Question

If tan² θ = 1 - e², show that sec θ + tan³ θ . cosec θ = $(2 - e^{2})^{\frac{3}{2}}$

Answer 1

Proof Using Trigonometric Identities

We are given a question of proof which we are supposed to solve by using Trigonometric Relations.

We will use the following two identities:

$\boxed{\begin{minipage}{10em} 1+\tan^2\theta=\sec^2\theta\\\\ \sin^2\theta+\cos^2\theta=1 \end{minipage}}$

We are given:

$\rightarrow \tan^2\theta=1-e^2\\\\\\ \textsf{Use } 1+\tan^2\theta=\sec^2\theta \\\\\\ \implies \sec^2\theta=1+(1-e^2)\\\\\\ \implies \sec^2\theta=2-e^2 \\\\\\\ \implies \sec\theta=(2-e^2)^{^1\!/_2} --- \mathbb{RESULT}$

Let us now take the LHS.

$\displaystyle\mathbb{LHS}\\\\\\=\sec\theta+\tan^3\theta.\,\text{cosec}\,\theta\\\\\\=\frac{1}{\cos\theta}+\frac{\sin^3\theta}{\cos^3\theta}.\frac{1}{\sin\theta}\\\\\\=\frac{1}{\cos\theta}+\frac{\sin^2\theta}{\cos^3\theta}\\\\\\=\frac{\cos^2\theta+\sin^2\theta}{\cos^3\theta}\\\\\\=\frac{1}{\cos^3\theta}\\\\\\=\sec^3\theta\\\\\\\textsf{Using the Result}\\\\\\=\left((2-e^2)^{^1\!/_2}\right)^3\\\\\\ = (2-e^2)^{^3\!/_2}\\\\\\ = \mathbb{RHS} \\\\\\ \mathcal{HENCE\ \ PROVED}$

Answer 2

HELLO DEAR,




GIVEN:- tan²θ = 1 - e²


we know:- 1 + tan²θ = sec²θ , sin²θ + cos²θ = 1


now, sec²θ = 1 + tan²θ

=> sec ²θ = 1 + (1 - e²)

=> sec²θ = 2 - e²

=> secθ = (2 - e²)½


so, secθ + tan³θ . cosecθ

=> 1/cosθ + sin³θ/cos³θ . 1/sinθ

=> (cos²θ + sin²)/cos³θ

=> 1/cos³θ

=> sec³θ = (2 - e²)$\bold{^{3/2}}$



I HOPE IT'S HELP YOU DEAR,
THANKS