Advertisements
Advertisements
प्रश्न
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
Advertisements
उत्तर
L.H.S = `(tan^2 theta - 1)/(tan^2 theta + 1)`
= `(tan^2 theta - 1)/(sec^2 theta)`
= `(sin^2 theta)/(cos^2 theta) - 1 ÷ 1/(cos^2 theta)`
= `(sin^2 theta - cos^2 theta)/(cos^2 theta) xx (cos^2 theta)/1`
= sin2θ − cos2θ
= 1 − cos2θ − cos2θ
= 1 – cos2θ
L.H.S = R.H.S
Hence it is proved.
APPEARS IN
संबंधित प्रश्न
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
Write the value of `(1 + tan^2 theta ) cos^2 theta`.
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
If sinA + cosA = `sqrt(2)` , prove that sinAcosA = `1/2`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
