Advertisements
Advertisements
प्रश्न
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
tan 65° + cot 49°
उत्तर
We know `tan (90^@ - theta) = cot theta` and `cot(90^@ - theta) = tan theta`. So
`tan 65° + cot 49° = tan(90^@ - 25^@) + cot 90^@ (90^@ - 41^@)`
`= cot 25^@ + tan 41^@`
Thus the desired expression is `= cot 25^@ + tan 41^@`
APPEARS IN
संबंधित प्रश्न
Evaluate the following:
`(sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)`
`(1- tan^2 45°)/(1+tan^2 45°)` = ______
Evaluate: `(3 cos 55^@)/(7 sin 35^@) - (4(cos 70 cosec 20^@))/(7(tan 5^@ tan 25^@ tan 45^@ tan 65^@ tan 85^@))`
If A = 30°;
show that:
sin 3 A = 4 sin A sin (60° - A) sin (60° + A)
If tan (A + B) = 1 and tan(A-B)`=1/sqrt3` , 0° < A + B < 90°, A > B, then find the values of A and B.
Prove that:
cos2 30° - sin2 30° = cos 60°
If A = 30o, then prove that :
2 cos2 A - 1 = 1 - 2 sin2A
Find the value of x in the following: 2 sin3x = `sqrt(3)`
If A = 30° and B = 60°, verify that: `(sin("A" -"B"))/(sin"A" . sin"B")` = cotB - cotA
Find the value of x if `2 "cosec"^2 30 + x sin^2 60 - 3/4 tan^2 30` = 10
