Advertisements
Advertisements
प्रश्न
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
Advertisements
उत्तर
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(cos^2θ + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ sin2θ cos2θ = 1
= `1/sinθ xx 1/cosθ`
= cosecθ × secθ
∴ L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
If`( 2 sin theta + 3 cos theta) =2 , " prove that " (3 sin theta - 2 cos theta) = +- 3.`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
From the figure find the value of sinθ.
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities.
`(sin^3"A" + cos^3"A")/(sin"A" + cos"A") + (sin^3"A" - cos^3"A")/(sin"A" - cos"A")` = 2
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt(a^2 + b^2 - c^2)`
Prove that `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`.
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
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`
(1 – cos2 A) is equal to ______.
