Question

Prove that
$(1 + \tan ^{2} a) + (1 + \frac{1}{ \tan ^{2}a } ) = \frac{1}{ { \sin }^{2}a - { \tan }^{4}a }$
​

Answer 1

Correct Question:

Prove that

$\displaystyle{\sf\:(\:1\:+\:\tan^2\:a\:)\:+\:\left(\:1\:+\:\dfrac{1}{\tan^2\:a}\:\right)\:=\:\dfrac{1}{\sin^2\:a\:-\:\bf{\sin^4\:a}}}$

Answer:

$\displaystyle{\boxed{\red{\sf\:(\:1\:+\:\tan^2\:a\:)\:+\:\left(\:1\:+\:\dfrac{1}{\tan^2\:a}\:\right)\:=\:\dfrac{1}{\sin^2\:a\:-\:\sin^4\:a}\:}}}$

Step-by-step-explanation:

We have given a trigonometric equation.

We have to prove that trigonometric equation.

The given trigonometric equation is

$\displaystyle{\sf\:(\:1\:+\:\tan^2\:a\:)\:+\:\left(\:1\:+\:\dfrac{1}{\tan^2\:a}\:\right)\:=\:\dfrac{1}{\sin^2\:a\:-\:\sin^4\:a}}$

$\displaystyle{\sf\:LHS\:=\:(\:1\:+\:\tan^2\:a\:)\:+\:\left(\:1\:+\:\dfrac{1}{\tan^2\:a}\:\right)}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:1\:+\:\tan^2\:a\:=\:\sec^2\:a\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sec^2\:a\:+\:\left(\:1\:+\:\dfrac{1}{\tan^2\:a}\:\right)}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\cot\:a\:=\:\dfrac{1}{\tan\:a}\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sec^2\:a\:+\:(\:1\:+\:\cot^2\:a\:)}$

We know that,

$\displaystyle{\boxed{\green{\sf\:1\:+\:\cot^2\:a\:=\:cosec^2\:a\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\sec^2\:a\:+\:cosec^2\:a}$

We know that,

$\displaystyle{\boxed{\purple{\sf\:\sec\:a\:=\:\dfrac{1}{\cos\:a}\:}}}$

$\displaystyle{\boxed{\orange{\sf\:cosec\:a\:=\:\dfrac{1}{\sin\:a}\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1}{\cos^2\:a}\:+\:\dfrac{1}{\sin^2\:a}}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\cos^2\:a\:=\:1\:-\:\sin^2\:a\:}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1}{1\:-\:\sin^2\:a}\:+\:\dfrac{1}{\sin^2\:a}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{\cancel{\sin^2\:a\:}\:+\:1\:-\:\cancel{\sin^2\:a}}{\sin^2\:a\:(\:1\:-\:\sin^2\:a\:)}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{1}{\sin^2\:a\:-\:\sin^4\:a}}$

$\displaystyle{\sf\:RHS\:=\:\dfrac{1}{\sin^2\:a\:-\:\sin^4\:a}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!

Answer 2

Question:-

Prove that:

${\sf \sf \large\:(\:1+tan^2\:A\:)\:+\:\left(\:1\:+\:\dfrac{1}{tan^2\:A}\:\right)\:=\:\dfrac{1}{sin^2\:A\:-\:{sin^4\:A}}}$

Given:-

${\sf \sf \large\:(\:1+tan^2\:A\:)\:+\:\left(\:1\:+\:\dfrac{1}{tan^2\:A}\:\right)\:=\:\dfrac{1}{sin^2\:A\:-\:{sin^4\:A}}}$

Solution:-

$\sf \large \: Identity : sec {}^{2} A−tan {}^{2} A=1$

$\sf \large L.HS. = (1 + tan {}^{2} A)(1 + \frac{1}{tan {}^{2}A } )$

$\sf \large = (sec {}^{2} A) \frac{sec {}^{2}A }{tan {}^{2}A }$

$\sf \large = \frac{1}{cos {}^{2} A} \: \frac{1}{ \frac{cos {}^{2}A }{ \frac{sin {}^{2}A }{cos {}^{2}A } } }$

$\sf \large = \frac{1}{sin {}^{2} A \: cos {}^{2} A}$

$\sf \large = \frac{1}{sin {}^{2} A(1 - sin {}^{2}A) }$

$\sf \large = \frac{1}{sin {}^{2} A - sin {}^{4} A}$

Answer:-

$\sf \: { \boxed{ \sf \red{Hence, Proved \: that = L.H.S. = R.H.S.}}} \\ \\ \sf \large{ \boxed{ \sf \green{(1 + tan {}^{2} A) + \bigg(1 + \frac{1}{tan {}^{2}A } \bigg) = \frac{1}{sin {}^{2} A - sin {}^{4} A} .}}}$

Hope you have satisfied. ⚘