Advertisements
Advertisements
Question
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Advertisements
Solution
`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
RELATED QUESTIONS
Evaluate cosec 31° − sec 59°
Show that cos 38° cos 52° − sin 38° sin 52° = 0
solve.
cos240° + cos250°
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
Find the value of x, if cos (2x – 6) = cos2 30° – cos2 60°
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
If \[\tan A = \frac{5}{12}\] \[\tan A = \frac{5}{12}\] find the value of (sin A + cos A) sec A.
The value of tan 10° tan 15° tan 75° tan 80° is
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))`
The value of tan 72° tan 18° is
