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प्रश्न
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
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उत्तर
LHS = sin (90° - θ) cos (90° - θ)
LHS = cos θ. sin θ
RHS = tan θ. cos2θ
RHS = `sin θ/cos θ` x cos2θ
RHS = cos θ. sin θ
∴ LHS = RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
Find the value of sin2θ + cos2θ
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`
