Question

(sin theta+ sec theta)^2 + (cos theta + cosectheta )^2 = (1 +
sec theta cosec theta)^2
​

Answer 1

To Prove

(sin∅ + sec∅)² + (cos∅ + cosec∅)² = (1 + sec∅cosec∅)²

$\rule{200}2$

Proof

$\rule{50}1$

Taking LHS

(sin∅ + sec∅)² + (cos∅ + cosec∅)²

= sin²∅ + sec²∅ + cos²∅ + cosec²∅ + 2sin∅sec∅ + 2cos∅cosec∅

[using, (a + b)² = a² + b² + 2ab]

= 1 + sec²∅ + cosec²∅ + 2sin∅sec∅ + 2cos∅cosec∅

(using the identity cos²∅ + sin²∅ = 1)

= 1 + $\sf\frac{1}{cos^2\theta} + \frac{1}{sin^2\theta} + \frac{2sin\theta}{cos\theta} + \frac{2cos\theta}{sin\theta}$

(using, sec∅ = 1/cos∅ and cosec∅ = 1/sin∅)

= 1 + $\sf\frac{(cos^2\theta + sin^2\theta)}{cos^2\theta sin^2 \theta} + \frac{(2sin^2\theta + 2cos^2\theta)}{sin\theta cos\theta}$

(Taking LCM and solving)

= 1 + $\sf\frac{1}{sin^2\theta cos^2\theta} + \frac{2(sin^2\theta + cos^2\theta)}{sin\theta cos\theta}$

(since, cos²∅ + sin²∅ = 1)

= 1 + sec²∅cosec²∅ + 2/sin∅cos∅

(using, 1/sin∅ = cosec∅ and 1/cos∅ = sec∅)

= 1 + sec²∅cosec²∅ + 2sec∅cosec∅

= (1 + sec∅cosec∅)²

= RHS

[using, a² + b² + 2ab = (a + b)²]

Hence Proved.

Answer 2

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Solution