Question

prove that tan^4A + tan^2A = SEC^4 A - sec^2A

Answer 1

Solution:

Given

LHS = tan⁴A+tan²A

= (tan²A)²+tan²A

= tan²A(tan²A+1)

= tan²Asec²A

________________________

*/ By Trigonometric identity:

1+tan²A = sec²A

Or

tan²A = sec²A-1*/

________________________

= (sec²A-1)sec²A

= sec⁴A-sec²A

= RHS

Therefore,

tan⁴A+tan²A = sec⁴A-sec²A

•••••

Answer 2

$\bold{\tan ^{4} A+\tan ^{2} A=\sec ^{2} A-\sec ^{2} A}$  is proven.

Solution:

The given equation comes under trigonometric identities where equality should be defined in both sides.

Given: $\tan ^{4} A+\tan ^{2} A=\sec ^{2} A-\sec ^{2} A$

Let us take L.H.S to prove the theorem

In L.H.S taking, $\tan ^{2} A$ common we get  

$\begin{aligned} \tan ^{2} A\left(\tan ^{2} A+1\right) &=\tan ^{2} A\left(\sec ^{2} A\right)\left(\because \tan ^{2} A+1=\sec ^{2} A\right) \\ &=\tan ^{2} A\left(\sec ^{2} A\right) \end{aligned}$

As, $\left(\because \tan ^{2} A=\sec ^{2} A-1\right), substitute\ \left(\sec ^{2} A-1\right) instead\ of\ \tan ^{2} A$

We get, $\tan ^{2} A\left(\sec ^{2} A\right)=\sec ^{2} A\left(\sec ^{2} A-1\right)$

Multiply $\sec ^{2} A$ with the terms inside the bracket.

 $=\sec ^{4} A-\sec ^{2} A$

Hence, L.H.S = R.H.S