Advertisements
Advertisements
Question
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Advertisements
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 the following trigonometric identities.
(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
Prove the following identities:
`sin theta/((cot theta + "cosec" theta)) - sin theta/((cot theta - "cosec" theta)) = 2`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Prove that :(sinθ+cosecθ)2+(cosθ+ secθ)2 = 7 + tan2 θ+cot2 θ.
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
