Advertisements
Advertisements
प्रश्न
Evaluate the following :
`(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)`
उत्तर
We have to find: `(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)`
Since `sec 70^@/(cosec 20^@) + sin 59^@/cos 31^@` and `sec (90^@ - theta) = cosec theta
So
`sec 70^@/(cosec 20^@) + sin 59^@/cos 31^@ = (sec (90^@ - 20^@))/(cosec 20^@) + (sin (90^@ - 31^@))/(cos 31^@)`
`= (cosec 20^@)/(cosec 20^@) + (cos 31^@)/(cos 31^@)`
1 + 1
= 2
So value of `(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)` is 2
APPEARS IN
संबंधित प्रश्न
An equilateral triangle is inscribed in a circle of radius 6 cm. Find its side.
sin 2A = 2 sin A is true when A = ______.
If Sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A =?
find the value of: sin2 30° + cos2 30°+ cot2 45°
ABC is an isosceles right-angled triangle. Assuming of AB = BC = x, find the value of each of the following trigonometric ratio: cos 45°
Given A = 60° and B = 30°,
prove that : cos (A + B) = cos A cos B - sin A sin B
Given A = 60° and B = 30°,
prove that : cos (A - B) = cos A cos B + sin A sin B
If A =30o, then prove that :
cos 2A = cos2A - sin2A = `(1 – tan^2"A")/(1+ tan^2"A")`
Find the value of x in the following: `2sin x/(2)` = 1
Verify the following equalities:
cos 90° = 1 – 2sin2 45° = 2cos2 45° – 1