Sum

Prove that (sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2} = 7 + tan^{2 }θ + cot^{2 }θ.

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#### Solution 1

L.H.S = (sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2}

= (sin^{2}θ + cosec^{2}θ + 2 sin θ cosec θ + cos^{2}θ + sec^{2}θ + 2cos θ sec θ)

= (sin^{2}θ + cos^{2}θ) + (cosec^{2}θ + sec^{2}θ) + 2 sin θ `(1/("sin"theta)) + 2 cos theta (1/("cos" theta))`

= (1) + (1 + cot^{2}θ + 1 + tan^{2}θ) + (2) + (2)

= 7 + tan^{2}θ + cot^{2}θ

= R.H.S

#### Solution 2

L.H.S = (sin θ + cosec θ)^{2} + (cos θ + sec θ)^{2}

= (sin^{2}θ + cosec^{2}θ + 2 sin θ cosec θ + cos^{2}θ + sec^{2}θ + 2cos θ sec θ)

= (sin^{2}θ + cos^{2}θ ) + 1 + cot^{2}θ + 2 sin θ x `1/sin θ` + 1 + tan^{2} θ + 2cos θ. `1/cos θ`

= 1 + 1 + 1 + 2 + 2 + tan^{2} θ + cot^{2}θ

= 7 + tan^{2} θ + cot^{2}θ

= RHS

Hence proved.

Concept: Trigonometric Identities

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