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प्रश्न
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
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उत्तर
Here,
x2 = a2 sec2θ + 2ab sec θ.tan θ + b2tan2θ
y2 = a2 tan2θ + 2ab sec θ.tan θ + b2sec2θ
⇒ x2 - y2 = a2 ( sec2θ - tan2θ ) - b2 ( sec2θ - tan2θ )
⇒ x2 - y2 = a2 - b2. ....( ∵ sec2θ - tan2θ = 1)
Hence proved.
संबंधित प्रश्न
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`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
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If cosA + cos2A = 1, then sin2A + sin4A = 1.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
