Advertisements
Advertisements
प्रश्न
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Advertisements
उत्तर
LHS = cos θ sin (90° - θ) + sin θ cos (90° - θ)
= cos θ. cos θ + sin θ. sin θ
= cos2θ + sin 2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
