Question

If tan theta minus cot theta equal to a and cos theta + sin theta equal to b prove that a square + 4 into b square minus 1 whole square equal to 4

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Given,

$tan \theta - cot \theta = a$ and $cos \theta +sin\theta = b$

Required to prove

(a² + 4) (b² - 1)² = 4

Recall the formula

$tan\theta = \frac{1}{cot\theta}$

$sin^2\theta+ cos^2\theta = 1$

Solution:

We have a = $tan \theta - cot \theta$

a² = $(tan \theta - cot \theta)^2$

= $tan^2\theta +cot^2\theta -2tan\theta cot\theta$

=$tan^2\theta +cot^2\theta -2$ (since $tan\theta = \frac{1}{cot\theta}$)

a² + 4 = $tan^2\theta +cot^2\theta +2$

Again we have

b = $cos\theta +sin\theta$

b² = $(cos\theta+sin\theta)^2$

= $cos^2\theta+sin^2\theta+2sin\theta cos\theta$ ( since $sin^2\theta+ cos^2\theta = 1$)

= $1+2sin\theta cos\theta$

b² - 1 = $2sin\theta \cos\theta$

(b² - 1)² = $4sin^2\theta\cos^2\theta$

LHS = (a² + 4) (b² - 1)² = $(tan^2\theta +cot^2\theta +2)4sin^2\theta cos^2\theta$

= $(\frac{sin^2\theta}{cos^\theta} +\frac{cos^2\theta}{sin^\theta} +2)4sin^2\theta cos^2\theta$

= $4sin^4\theta+ 4cos^4\theta+8sin^2\theta cos^2\theta$

=$4(sin^4\theta+ 4cos^4\theta+2sin^2\theta cos^2\theta)$

=$4(sin^2\theta+ cos^2\theta)^2$(since$sin^2\theta+ cos^2\theta = 1$)

=4

= RHS

(a² + 4) (b² - 1)² = 4

Hence proved

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