Advertisements
Advertisements
प्रश्न
Prove that `(sin ("B - C"))/(cos "B" cos "C") + (sin ("C - A"))/(cos "C" cos "A") + (sin ("A - B"))/(cos "A" cos "B")` = 0
Advertisements
उत्तर
Consider `(sin ("B - C"))/(cos "B" cos "C")`
`= (sin "B" cos "C" - cos "B" sin "C")/(cos "B" cos "C")`
`= (sin "B" cos "C")/(cos "B" cos "C") - (cos "B" sin "C")/(cos "B" cos "C")`
= tan B – tan C ……… (1)
Similarly we can prove `(sin ("C - A"))/(cos "C" cos "A")` = tan C – tan A …….(2)
and `(sin ("A - B"))/(cos "A" cos "B")` = tan A – tan B …….. (3)
Add (1), (2) and (3) we get
`(sin ("B - C"))/(cos "B" cos "C") + (sin ("C - A"))/(cos "C" cos "A") + (sin ("A - B"))/(cos "A" cos "B")` = 0
APPEARS IN
संबंधित प्रश्न
Find the value of the following:
cosec 15º
Find the value of the following:
sin (-105°)
If sin A = `3/5`, 0 < A < `pi/2` and cos B = `(-12)/13`, π < B < `(3pi)/2`, find the values of the following:
- cos(A + B)
- sin(A – B)
- tan(A – B)
Prove that:
sin(A + 60°) + sin(A – 60°) = sin A.
If sin A = `3/5`, find the values of cos 3A and tan 3A.
Find the value of tan `pi/8`.
Show that `cos^-1 (12/13) + sin^-1 (3/5) = sin^-1 (56/65)`
The value of cos(-480°) is:
If sin A + cos A = 1 then sin 2A is equal to:
The value of `(2 tan 30^circ)/(1 + tan^2 30^circ)` is:
