Advertisements
Advertisements
प्रश्न
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
उत्तर
`3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`
= `3(sin(90° - 18°))/(cos 18°) - sec(90° - 58°)/("cosec"58°)`
= `3(cos 18°)/(cos 18°) - ("cosec"58°)/("cosec"58°)`
= 3 - 1
= 2
APPEARS IN
संबंधित प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Evaluate.
sin(90° - A) cosA + cos(90° - A) sinA
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
Find the value of x, if tan x = `(tan60^circ - tan30^circ)/(1 + tan60^circ tan30^circ)`
If \[\tan A = \frac{3}{4} \text{ and } A + B = 90°\] then what is the value of cot B?
If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
If x tan 60° cos 60°= sin 60° cot 60°, then x = ______.
