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प्रश्न
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
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उत्तर
LHS = sin( 90° - θ ) sin θ cot θ
= cos θ . sin θ . `cos θ/sin θ`
= cos2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Write 'True' or 'False' and justify your answer in the following:
\[\cos \theta = \frac{a^2 + b^2}{2ab}\], where a and b are two distinct numbers such that ab > 0.
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
If x = acosθ , y = bcotθ , prove that `a^2/x^2 - b^2/y^2 = 1.`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 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`
