Advertisements
Advertisements
प्रश्न
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Advertisements
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
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Prove that:
sec (70° – θ) = cosec (20° + θ)
If the angle θ = –45° , find the value of tan θ.
Write the value of tan 10° tan 15° tan 75° tan 80°?
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
The value of cos2 17° − sin2 73° is
If \[\frac{x {cosec}^2 30°\sec^2 45°}{8 \cos^2 45° \sin^2 60°} = \tan^2 60° - \tan^2 30°\]
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
Evaluate: `(sin 80°)/(cos 10°)`+ sin 59° sec 31°
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
