Advertisements
Advertisements
Question
Find the value of `θ(0^circ < θ < 90^circ)` if :
`cos 63^circ sec(90^circ - θ) = 1`
Solution
`cos63^circ sec(90^circ - θ) = 1`
`cos 63^circ cosecθ = 1`
⇒ `cos63^circ = sinθ`
⇒ `cos 63^circ = cos(90^circ - θ)`
⇒ `63^circ = 90^circ - θ`
⇒ `θ = 27^circ`
APPEARS IN
RELATED QUESTIONS
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`
`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`
cosec4θ − cosec2θ = cot4θ + cot2θ
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove the following identity :
`(cot^2θ(secθ - 1))/((1 + sinθ)) = sec^2θ((1-sinθ)/(1 + secθ))`
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
