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Prove that (Sin θ + Cosec θ)2 + (Cos θ + Sec θ)2 = 7 + Tan2 θ + Cot2 θ. - Mathematics

Sum

Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tanθ + cotθ. 

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Solution 1

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

= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)

= (sin2θ + cos2θ) + (cosec2θ + sec2θ) + 2 sin θ `(1/("sin"theta)) + 2 cos theta (1/("cos" theta))`

= (1) + (1 + cot2θ + 1 + tan2θ) + (2) + (2)

= 7 + tan2θ + cot2θ 

= R.H.S

Solution 2

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

= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)

= (sin2θ + cos2θ ) + 1 + cot2θ + 2 sin θ x `1/sin θ` + 1 + tan2 θ + 2cos θ. `1/cos θ`

= 1 + 1 + 1 + 2 + 2 + tan2 θ + cot2θ

= 7 + tan2 θ + cot2θ

= RHS

Hence proved.

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APPEARS IN

ICSE Class 10 Mathematics
Chapter 18 Trigonometry
Exercise | Q 64.4
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