Advertisements
Advertisements
प्रश्न
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
उत्तर
`1-sin^2A/(1 + cosA)`
= `(1 + cosA - sin^2A)/(1 + cosA)`
= `(cosA + cos^2A)/(1 + cosA)`
= `(cosA(1 + cosA))/(1 + cosA)`
= cos A
APPEARS IN
संबंधित प्रश्न
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
`(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))`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.