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प्रश्न
If cot θ =` 7/8` evaluate `((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))`
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उत्तर १
Let us consider a right triangle ABC, right-angled at point B.
cot theta = `7/8`
If BC is 7k, then AB will be 8k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
= (8k)2 + (7k)2
= 64k2 + 49k2
= 113k2
AC = `sqrt113k`
`sin theta = (8k)/sqrt(113k) = 8/sqrt(113)`
`cos theta = (7k)/sqrt(113k) = 7/sqrt113`
`((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ)) = (1-sin^2θ)/(1-cos^2θ)`
= `(1-(8/sqrt113)^2)/(1-(7/sqrt(113))^2)`
= `(1-64/113) /(1-49/113)`
= `(49/113)/(64/113)`
= `49/64`
उत्तर २
`cot theta = 7/8`
`((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))`
= `(1 - sin^2 theta)/(1 - cos^2 theta)` ...[∵ (a + b) (a – b) = a2 − b2] a = 1, b = sin 𝜃
We know that sin 2𝜃 + cos2𝜃 = 1
1 − sin2𝜃 = cos2𝜃 = cos2𝜃
1 − cos2𝜃 = sin2 𝜃
= `(cos^2 theta)/(sin^2 theta)`
= `cot^2 theta`
= `(cot theta)^2`
= `[7/8]^2`
= `49/64`
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