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प्रश्न
If x = a cos θ and y = b cot θ, show that:
`a^2/x^2 - b^2/y^2 = 1`
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उत्तर
`a^2/x^2 - b^2/y^2`
= `a^2/(a^2cos^2theta) - b^2/(b^2cot^2theta)`
= `1/cos^2theta - sin^2theta/cos^2theta`
= `(1 - sin^2theta)/cos^2theta`
= `cos^2theta/cos^2theta`
= 1
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संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
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`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
