Advertisements
Advertisements
Question
Prove that:
tan (55° - A) - cot (35° + A)
Advertisements
Solution
tan (55° - A) - cot (35° + A)
= tan [90° - (55° - A)] - cot (35° + A)
= cot (90° - 55° + A) - cot (35° + A)
= cot (35° + A) - cot (35° + A)
= 0
RELATED QUESTIONS
Evaluate.
cos225° + cos265° - tan245°
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Use tables to find sine of 62° 57'
If 16 cot x = 12, then \[\frac{\sin x - \cos x}{\sin x + \cos x}\]
If \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
A, B and C are interior angles of a triangle ABC. Show that
sin `(("B"+"C")/2) = cos "A"/2`
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.
Find the value of the following:
sin 21° 21′
The value of 3 sin 70° sec 20° + 2 sin 49° sec 51° is
