Advertisements
Advertisements
Question
Prove that:
`(sin(180^circ - theta)cos(90^circ + theta)tan(270^circ - theta)cot(360^circ - theta))/(sin(360^circ - theta)cos(360^circ + theta)sin(270^circ - theta)cosec (-theta))` = -1
Solution
LHS = `(sin(180^circ - theta)cos(90^circ + theta)tan(270^circ - theta)cot(360^circ - theta))/(sin(360^circ - theta)cos(360^circ + theta)sin(270^circ - theta)cosec (-theta))`
`= ((sin theta)(- sin theta)(cot theta)(- cot theta))/((- sin theta)(cos theta)(- cos theta)(- cosec theta))`
`= (- sin theta xx cot theta cot theta)/(cos theta xx cos theta "cosec" theta)`
`= (- sin theta xx (cos theta)/(sin theta) xx (cos theta)/(sin theta))/(cos theta xx cos theta xx 1/(sin theta))`
`= - 1 xx (sin theta)/(sin theta)` = - 1 = RHS
APPEARS IN
RELATED QUESTIONS
Find the degree measure corresponding to the following radian measure.
`(9pi)/5`
Find the degree measure corresponding to the following radian measure.
-3
Find the values of the following trigonometric ratio.
cos (-210°)
Find the values of the following trigonometric ratio.
cosec 1125°
Prove that:
2 sin2 `pi/6` + cosec2 `(7pi)/6` cos2 `pi/3 = 3/2`
Prove that:
`sin theta * cos theta {sin(pi/2 - theta) * "cosec" theta + cos (pi/2 - theta) * sec theta}` = 1
Prove that:
`(sin(180^circ + "A")cos(90^circ - "A")tan(270^circ - "A"))/(sec(540^circ - "A") cos(360^circ + "A") "cosec"(270^circ + "A"))` = - sin A cos2 A.
The degree measure of `pi/8` is
The value of `1/("cosec" (-45^circ))` is:
`((cos x)/(cosec x)) - sqrt(1 - sin^2x) sqrt(1 - cos^2 x)` is: