Advertisements
Advertisements
प्रश्न
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
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.
APPEARS IN
संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
Write the value of tan1° tan 2° ........ tan 89° .
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Find the value of sin 30° + cos 60°.
Prove that
sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A