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Question
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
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Solution
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.
RELATED QUESTIONS
Prove that:
2 sin2 A + cos4 A = 1 + sin4 A
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Prove the following identity :
sinθcotθ + sinθcosecθ = 1 + cosθ
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
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
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
