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prove that (sinθ + cosecθ)^2 + (cosθ + secθ)^2=7+tan^2θ+cot^2θ

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mun55 | Student, Grade 10 | eNoter

Posted September 23, 2012 at 4:47 PM via web

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prove that (sinθ + cosecθ)^2 + (cosθ + secθ)^2=7+tan^2θ+cot^2θ

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 23, 2012 at 5:15 PM (Answer #1)

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`(sintheta+cosectheta)^2+(costheta+sectheta)^2`

`=sin^2theta+cosec^2theta+2sintheta*costheta+cos^2theta+sec^2theta+2costheta*sectheta`

`= sin^2theta+cos^2theta+2+2+1+cot^2theta+1+tan^2theta`

`= 1+2+2+1+1+tan^2theta+cot^2theta`

`= 7+tan^2theta+cot^2theta`

Sources:

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 23, 2012 at 5:25 PM (Answer #2)

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The relation `(sin x + cosec x)^2 + (cos x + sec x)^2 = 7 + tan^2 x + cot^2x` has to be proved. (` theta` has been substituted with x)

Start with the left hand side:

`(sin x + cosec x)^2 + (cos x + sec x)^2`

=> `sin^2 x + cosec^2x + 2*sin x*cosec x + cos^2x + sec^2x + 2*cos x*sec x`

=> `1 + cosec^2x + 2 + sec^2x + 2`

=> `1 + 1 + cot^2x + 2 + 1 + tan^2x + 2`

=> `7 + cot^2x + tan^2x`

This proves that `(sin x + cosec x)^2 + (cos x + sec x)^2 = 7 + tan^2 x + cot^2x`

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