Advertisements
Advertisements
प्रश्न
Solve the following equation: `(cos θ)/(1 - sin θ) + cos θ/(1 + sinθ) = 4`.
उत्तर
We have,
∴ `(cos θ)/(1 - sin θ) + (cos θ)/(1 + sin θ) = 4`
`⇒ cos θ{(1)/(1 - sin θ) + (1)/(1 + sin θ)} = 4`
`⇒ cos θ{(1 + sin θ + 1 - sin θ)/((1 - sin θ)(1 + sin θ))} = 4`
⇒ 2cos θ = 4( 1 - sinθ )( 1 + sin θ)
⇒ 2cos θ = 4( 1 - sin2θ )
⇒ 2 cos θ = 4cos2θ
⇒ 4cos2θ - 2cos θ = 0
⇒ 2cos θ( 2cos θ - 1) = 0
⇒ 2cos θ = 0 or ⇒ 2cos θ - 1 = 0
⇒ cos θ = 0 or ⇒ cos θ = `1/2`
⇒ θ = 60°, (since 0 < θ < 90° ).
APPEARS IN
संबंधित प्रश्न
Evaluate without using trigonometric tables,
`sin^2 28^@ + sin^2 62^@ + tan^2 38^@ - cot^2 52^@ + 1/4 sec^2 30^@`
Without using trigonometric tables, prove that:
sin248° + sin242° = 1
Without using trigonometric tables, prove that:
(sin72° + cos18°)(sin72° − cos18°) = 0
Prove that:
`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`
Prove that:
sin θ cos (90° - θ ) + sin (90° - θ) cos θ = 1
Prove that:
\[\frac{\sin\theta}{\cos(90° - \theta)} + \frac{\cos\theta}{\sin(90° - \theta)} = 2\]
Prove the following:
`1/(1+sin^2theta) + 1/(1+cos^2theta) + 1/(1+sec^2theta) + 1/(1+cosec^2theta) = 2`
The maximum value of `1/(cosec alpha)` is ______.
Prove that:
`(cos^2 "A")/(cos "A" - sin "A") + (sin "A")/(1 - cot "A")` = sin A + cos A
Prove that:
`(sin^3 theta + cos^3 theta)/(sin theta + cos theta) = 1 - sin theta cos theta`
