Advertisements
Advertisements
प्रश्न
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
Prove the following:
cosec4 θ − cosec2 θ = cot2 θ + cot4 θ
Advertisements
उत्तर १
LHS = cosec4 θ − cosec2 θ
LHS = cosec2 θ (cosec2 θ − 1)
LHS = (cot2 θ + 1)cot2 θ ...`{(cot^2 θ + 1 = cosec^2 θ),(∵ cot^2 θ = cosec^2 θ - 1):}`
LHS = cot4 θ + cot2 θ
RHS = cot4 θ + cot2 θ
RHS = LHS
Hence proved.
उत्तर २
RHS = cot4 θ + cot2 θ
RHS = cot2 θ (cot2 θ + 1)
RHS = (cosec2 θ − 1)cosec2 θ ...`{(cot^2 θ + 1=cosec^2 θ),(∵ cot^2θ = cosec^2 θ - 1):}`
RHS = cosec4 θ − cosec2 θ
LHS = cosec4 θ − cosec2 θ
RHS = LHS
Hence proved.
संबंधित प्रश्न
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
