Advertisements
Advertisements
प्रश्न
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Advertisements
उत्तर
cosec68° + cot72°
= cosec(90 - 22)° + cot(90 -18)°
= sec22° + tan18°
संबंधित प्रश्न
Evaluate `(sin 18^@)/(cos 72^@)`
if `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
Evaluate:
cosec (65° + A) – sec (25° – A)
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
If 3 cot θ = 4, find the value of \[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\]
If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Find the value of the following:
`((cos 47^circ)/(sin 43^circ))^2 + ((sin 72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
