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प्रश्न
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
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उत्तर
L.H.S. = `1/(tan A + cot A)`
= `1/((sin A)/(cos A) + (cos A)/(sin A))`
= `1/((sin^2A + cos^2A)/(sin A cos A))`
= `1/(1/(sin A cos A))` ...(∵ sin2A + cos2A = 1)
= sin A cos A
= R.H.S.
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
