Advertisements
Advertisements
Question
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Advertisements
Solution
L.H.S. = `(sinthetasin(90^@-theta))/cot(90^@-theta)`
= `(sinthetacostheta)/tantheta`
= `(sinthetacostheta)/(sintheta/costheta)`
= cos2θ
= 1 – sin2θ = R.H.S.
APPEARS IN
RELATED QUESTIONS
If the angle θ = -60° , find the value of sinθ .
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Solve.
sin15° cos75° + cos15° sin75°
Solve.
sin42° sin48° - cos42° cos48°
Evaluate.
`(2tan53^@)/(cot37^@)-cot80^@/tan10^@`
For triangle ABC, show that: `sin (A + B)/2 = cos C/2`
Evaluate:
`sec26^@ sin64^@ + (cosec33^@)/sec57^@`
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
