Advertisements
Advertisements
प्रश्न
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
पर्याय
\[\frac{x^2 + 1}{x}\]
\[\frac{x^2 - 1}{x}\]
\[\frac{x^2 + 1}{2x}\]
\[\frac{x^2 - 1}{2x}\]
Advertisements
उत्तर
Given:
`sec θ+tanθ=x`
We know that,
`sec^2 θ-tan^2 θ=1`
⇒` (sec θ+tan θ)(sec θ-tanθ)=1`
⇒`x(sec θ-tan θ)=1`
⇒ `secθ-tan θ=1/x`
Now,
`secθ+tan θ=x,`
`sec θ-tan θ=1/x`
Subtracting the second equation from the first equation, we get
`(secθ+tan θ)-(secθ-tanθ)=x-1/x`
⇒` secθ+tanθ-secθ+tanθ=(x^2-1)/x`
⇒ `2 tanθ=(x^2-1)/x`
⇒ `2 tan θ=(x^2-1)/(2x)`
⇒ `tan θ=(x^2-1)/(2x)`
APPEARS IN
संबंधित प्रश्न
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Prove the following identity :
`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
If 1 – cos2θ = `1/4`, then θ = ?
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
