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Question
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
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Solution
LHS = sec2 θ + cosec2 θ
= `1/(cos^2 θ) + 1/(sin^2 θ)`
= `(sin^2 θ + cos^2 θ)/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ. cos^2 θ)`
= `1/(sin^2 θ) xx 1/(cos^2 θ)`
= sec2 θ cosec2 θ
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`
`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle.
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
(1 + sin A)(1 – sin A) is equal to ______.
