Advertisements
Advertisements
Question
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
Advertisements
Solution
LHS = cosec2(90° - θ) - tan2 θ
LHS = sec2 θ - tan2 θ = 1
RHS = cos2(90° - θ) + cos2 θ
RHS = sin2θ + cos2 θ = 1
Hence, LHS = RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
