Advertisements
Advertisements
प्रश्न
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove the following:
`(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Advertisements
उत्तर
LHS = `(sin θ/cos θ + sin θ)/(sin θ/cos θ - sin θ)`
= `(sin θ (1/cos θ + 1))/(sin θ (1/cos θ - 1))`
= `(sec θ + 1)/(sec θ - 1)`
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities:
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
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`
