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Question
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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Solution
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.
RELATED QUESTIONS
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
If x cos A + y sin A = m and x sin A – y cos A = n, then prove that : x2 + y2 = m2 + n2
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
(sec A + tan A) (1 − sin A) = ______.
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
