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प्रश्न
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
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उत्तर
L.H.S:
`tanA/(1 + sec A) - tanA/(1 - sec A)`
Taking LCM of the denominators,
= `(tanA(1 - sec A) - tanA(1 + sec A))/((1 + sec A)(1 - sec A))`
Since, (1 + sec A)(1 – sec A) = 1 – sec2A
= `(tan A(1 - secA - 1 - sec A))/(1 - sec^2A)`
= `(tan A(-2 sec A))/(1 - sec^2 A)`
= `(2 tan A *sec A)/(sec^2 A - 1)`
Since,
sec2A – tan2A = 1
sec2A – 1 = tan2A
= `(2 tan A * sec A)/(tan^2 A)`
Since, sec A = `(1/cosA)` and tan A = `(sinA/cosA)`
= `(2secA)/tanA = (2cosA)/(cosA sinA)`
= `2/sinA`
= 2 cosec A ...`(∵ 1/sinA = "cosec" A)`
= R.H.S
Hence proved.
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Prove the following identities:
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Prove the following identity :
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Prove the following identity :
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Prove the following identity :
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a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
(sec θ + tan θ) . (sec 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`
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
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]
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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`
