Advertisements
Advertisements
प्रश्न
Evaluate the following :
(sin 72° + cos 18°) (sin 72° − cos 18°)
उत्तर
We have to find: (sin 72° + cos 18°) (sin 72° − cos 18°)
Since `sin(90^@ - theta) = cos theta` So
sin 72° + cos 18°) (sin 72° − cos 18°) = `(sin 72^@)^2 - (cos 18^@)^2`
`= [sin (90^@ - 18^@)]^2 - (cos 18^@)^2`
`= (cos 18^@)^2 - (cos 18^@)^2`
`= cos^2 18^@ - cos^2 18^@`
= 0
So value of `(sin 72^@ + cos 18^@)(sin 72^@ - cos 18^@)` is 0
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
`(5cos^2 60° + 4sec^2 30° - tan^2 45°)/(sin^2 30° + cos^2 30°)`
Evaluate: tan 7° tan 23° tan 60° tan 67° tan 83°
If sin x = cos y, then x + y = 45° ; write true of false
If A =30o, then prove that :
sin 2A = 2sin A cos A = `(2 tan"A")/(1 + tan^2"A")`
Without using table, find the value of the following:
`(sin30° - sin90° + 2cos0°)/(tan30° tan60°)`
Find the value of x in the following: tan x = sin45° cos45° + sin30°
Find the value of the following:
`(tan45^circ)/("cosec"30^circ) + (sec60^circ)/(cot45^circ) - (5sin90^circ)/(2cos0^circ)`
If sin 30° = x and cos 60° = y, then x2 + y2 is
If A and B are acute angles such that sin (A – B) = 0 and 2 cos (A + B) – 1 = 0, then find angles A and B.
If sin α = `1/2`, then find the value of (3 cos α – 4 cos3 α).