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प्रश्न
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
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उत्तर
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
= `sin^2A + cosec^2A + 2sinA xx 1/sinA + cos^2A + sec^2A + 2cosA xx 1/cosA`
= sin2 A + cos2 A + cosec2 A + sec2 A + 2 + 2 ...(∵ sin2 A + cos2 A = 1)
= 1 + cosec2 A + sec2 A + 4
= (1 + cot2 A) + (1 + tan2 A) + 5 ...[∵ cosec2 A = 1 + cot2 A and sec2 A = 1 + tan2 A]
= 1 + cot2 A + 1 + tan2 A + 5
= 7 + tan2 A + cot2 A
= R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
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
