Question

If sin^4x+sin^2x = 1 then prove that tan^4x-tan^2x = 1

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$tan^4x -tan^2x = 1\ is\ proved$

Solution:

Given that,

$sin^4x + sin^2x = 1$

We have to prove that:

$tan^4x -tan^2x = 1$

From given,

$sin^4x + sin^2x = 1\\\\sin^4x = 1 - sin^2 x\\\\(1 - sin^2 x = cos^2 x )\\\\sin^4x = cos^2 x$

We can rewrite the above as:

$sin^2x \times sin^2 x = cos^2 x\\\\\frac{sin^2 x}{cos^2 x} = \frac{1}{sin^2 x}\\\\( \frac{sin x}{cos x} = tan x )\\\\Therefore\\\\tan^2 x = \frac{1}{sin^2 x}\\\\( \frac{1}{sinx} = cosecx)\\\\Therefore\\\\tan^2 x = cosec^2 x$

$(1+cot^2 x = cosec^2 x)\\\\Therefore\\\\tan^2x = 1+cot^2 x\\\\Multiply\ each\ term\ by\ tan^2x\\\\tan^4x = tan^2 x + cot^2 x tan^2 x\\\\tan^4x = tan^2 x + 1\\\\tan^4x - tan^2 x = 1$

Thus proved

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