Advertisements
Advertisements
प्रश्न
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Advertisements
उत्तर
LHS = cos 1°cos 2°cos 3° ....cos 180°
= cos 1°cos 2°cos 3° ....cos 89° cos 90° .... cos 180°
= cos 1°cos 2°cos 3° ....cos 89° x 0 x cos 91° .... cos 180°
= 0
= RHS
Hence proved.
संबंधित प्रश्न
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
If tan α + cot α = 2, then tan20α + cot20α = ______.
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
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
