Advertisements
Advertisements
Question
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
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= –60º, find the value of cosθ.
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
Solve.
`tan47/cot43`
Solve.
`sec75/(cosec15)`
Use tables to find cosine of 8° 12’
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If θ is an acute angle such that sec2 θ = 3, then the value of \[\frac{\tan^2 \theta - {cosec}^2 \theta}{\tan^2 \theta + {cosec}^2 \theta}\]
If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°), then tan θ is equal to
If sin 3A = cos 6A, then ∠A = ?
If sec A + tan A = x, then sec A = ______.
