Advertisements
Advertisements
प्रश्न
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
विकल्प
\[\frac{x^2 + 1}{x}\]
\[\frac{x^2 - 1}{x}\]
\[\frac{x^2 + 1}{2x}\]
\[\frac{x^2 - 1}{2x}\]
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: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
9 sec2 A − 9 tan2 A = ______.
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
`sin^2 theta + 1/((1+tan^2 theta))=1`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Write the value of tan10° tan 20° tan 70° tan 80° .
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove that: `(1 + cot^2 θ/(1 + cosec θ)) = cosec θ`.
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
