Question

prove that tan²A(1+cot²A)/(1+tan²A)=1​

Answer 1

Step-by-step explanation:

Given: tan²A(1+cot²A)/(1+tan²A)

Solving numerator:-

Tan=P/B

tan²=P²/B²

Cot=B/P

cot²=B²/P²

Simplifying:-

Tan²A(1+cot²A)=Tan²A+Tan²Cot²A²

Putting values:-

(P²/B²)•A + A²

or, A²+A•(P²/B²)

Solving denominator:-

1+tan²A= 1+(P²/B²)•A

Divide and prove it to 1 by taking same denomination.

The question should not include A in numerator, it is extra, perhaps, here is an idea for solving this question.

I have done according to 9th standard.

Answer 2

$\large\underline{\sf{Solution-}}$

Consider Left Hand Side

We have given that

$\rm :\longmapsto\:\dfrac{ {tan}^{2}A(1 + {cot}^{2} A)}{1 + {tan}^{2} A}$

From Trigonometric Identities, we have

$\rm :\longmapsto\:cotx = \dfrac{1}{tanx}$

So, on substituting this

$\rm \: = \: \: \dfrac{ {tan}^{2} A\bigg(1 + \dfrac{1}{ {tan}^{2} A} \bigg) }{1 + {tan}^{2} A}$

$\rm \: = \: \: \dfrac{ {tan}^{2} A\bigg(\dfrac{ {tan}^{2}A + 1}{ {tan}^{2} A} \bigg) }{1 + {tan}^{2} A}$

$\rm \: = \: \: 1$

Hence,

$\bf :\longmapsto\:\dfrac{ {tan}^{2}A(1 + {cot}^{2} A)}{1 + {tan}^{2} A} = 1$

Hence, Proved.

More to know

$\rm :\longmapsto\: {sin}^{2}x + {cos}^{2}x = 1$

$\rm :\longmapsto\: {cosec}^{2}x - {cot}^{2}x = 1$

$\rm :\longmapsto\: {sec}^{2}x - {tan}^{2}x = 1$