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प्रश्न
Prove the following identities:
`(1 + (secA - tanA)^2)/(cosecA(secA - tanA)) = 2tanA`
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उत्तर
`(1 + (secA - tanA)^2)/(cosecA(secA - tanA))`
= `((sec^2A - tan^2A) + (secA - tanA)^2)/(cosecA(secA - tanA))`
= `((secA - tanA)(secA + tanA) + (secA + tanA)^2)/(cosecA(secA - tanA))`
= `((secA + tanA) + (secA - tanA))/(cosecA)`
= `(2secA)/(cosecA)`
= `2(1/cosA)/(1/sinA)`
= 2 tanA
संबंधित प्रश्न
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
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 `"cot A"/(1 - tan "A") + "tan A"/(1 - cot"A")` = 1 + tan A + cot A = sec A . cosec A + 1
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]
cosec θ = `41/9`
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`
