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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 that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
If sin θ + cos θ = x, prove that `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
` tan^2 theta - 1/( cos^2 theta )=-1`
Evaluate:
`(tan 65°)/(cot 25°)`
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
