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प्रश्न
If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ
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उत्तर
sec θ = `41/40` ......[Given]
∴ cos θ = `1/sectheta = 1/(41/40)`
∴ cos θ = `40/41`
We know that,
sin2θ + cos2θ = 1
∴ `sin^2theta + (40/41)^2` = 1
∴ `sin^2theta + 1600/1681` = 1
∴ sin2θ = `1 - 1600/1681`
∴ sin2θ = `(1681- 1600)/1681`
∴ sin2θ = `81/1681`
∴ sin θ = `9/41` .......[Taking square root of both sides]
Now, cosec θ = `1/sintheta`
= `1/((9/41))`
= `41/9`
cot θ = `costheta/sintheta`
= `((40/41))/((9/41))`
= `40/9`
∴ sin θ = `9/41`, cot θ = `40/9`, cosec θ = `41/9`
संबंधित प्रश्न
Prove that: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
Prove the following identity :
`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
sin(45° + θ) – cos(45° – θ) is equal to ______.
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
