Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Convert the following degree measure into radian measure.
150°
Find the degree measure corresponding to the following radian measure.
`(11pi)/18`
Determine the quadrant in which the following degree lie.
380°
Find the values of the following trigonometric ratio.
sec 390°
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.
Prove that:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x)` = cot 3x
The value of `(3 tan 10^circ - tan^3 10^circ)/(1 - 3 tan^2 10^circ)` is:
The value of cosec-1 `(2/sqrt3)` is:
If α and β be between 0 and `pi/2` and if cos(α + β) = `12/13` and sin (α – β) = `3/5` then sin 2α is:
