Advertisements
Advertisements
प्रश्न
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.
उत्तर
LHS = `(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")(sin "A")(cot "A"))/((- sec "A")(cos "A")(- sec "A"))` .....`[(sec (540° - "A")),(= sec (360° + 180° - "A")),(= sec (180° - "A")),(= (- sec "A"))]`
`= (- sin "A" sin "A" (cos "A")/(sin "A"))/(- 1/(cos "A")cos "A" - 1/(cos "A"))`
= - sin A × cos A × cos A
= - sin A cos2A = RHS
APPEARS IN
संबंधित प्रश्न
Convert the following degree measure into radian measure.
60°
Find the degree measure corresponding to the following radian measure.
`pi/8`
Find the degree measure corresponding to the following radian measure.
`(9pi)/5`
Determine the quadrant in which the following degree lie.
380°
Prove that:
2 sin2 `pi/6` + cosec2 `(7pi)/6` cos2 `pi/3 = 3/2`
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
Prove that:
`sin theta * cos theta {sin(pi/2 - theta) * "cosec" theta + cos (pi/2 - theta) * sec theta}` = 1
If sin θ = `3/5`, tan φ = `1/2 and pi/2` < θ < π < φ < `(3pi)/2,`, then find the value of 8 tan θ – `sqrt5` sec φ.
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:
