Advertisements
Advertisements
प्रश्न
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
Advertisements
उत्तर
L.H.S. = `sinAcosA - (sinAsinAcosA)/(cosecA) - (cosAcosAsinA)/secA`
= sin A cos A – sin2 A cos A sin A – cos2 A sin A cos A
= sin A cos A – sin3 A cos A – cos3 A sin A
= sin A cos A [1 – sin2 A – cos2 A]
= sin A cos A [1 – (sin2 A + cos2 A)]
= sin A cos A (1 – 1)
= sin A cos A × 0
= 0 = R.H.S.
APPEARS IN
संबंधित प्रश्न
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
9 sec2 A − 9 tan2 A = ______.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove that `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`.
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
