Question

Prove that
(1+tan²A)+(1+1/tan²A)=1/sin²A-sin⁴A.​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

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

Solution :-

Taking L.H.S :-

$\left(1+tan^2A\right)+\left(1+\dfrac{1}{tan^2A}\right)$

$=\left(1+(tanA)^2\right)+\left(1+\bigg(\dfrac{1}{tan^2A}\bigg)^2\right)$

We know that "tanA = sinA/cosA"

$=\left(1+\bigg(\dfrac{sinA}{cosA}\bigg)^2\right)+\left(1+\left(\dfrac{1}{\dfrac{sinA}{cosA}}\right)^2\right)$

$= 1+\dfrac{sin^2A}{cos^2A}+1+\dfrac{cos^2A}{sin^2A}$

$= \dfrac{cos^2A+sin^2A}{cos^2A}+\dfrac{sin^2A+cos^2A}{sin^2A}$

We know that "sin^2A+cos^2A = 1"

$=\dfrac{1}{cos^2A}+\dfrac{1}{sin^2A}$

Taking L.C.M :-

$= \dfrac{sin^2A+cos^2A}{cos^2A\times sin^2A}$

We know that "sin^2A+cos^2A = 1"

$=\dfrac{1}{cos^2Asin^2A}$

From the trigonometric identity " sin^2A+cos^2A=1"

We can write "cos^2A = 1 - sin^2A"

$=\dfrac{1}{(1-sin^2A)sin^2A}$

$=\dfrac{1}{sin^2A-sin^4A}$

= R.H.S

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