Advertisements
Advertisements
प्रश्न
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Advertisements
उत्तर १
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
RHS = `(tanθ + secθ + 1)/(tanθ + secθ - 1)`
`"RHS" = (tanθ + secθ + 1)/((tanθ + secθ) - 1) × (tanθ + secθ + 1)/((tanθ + secθ) + 1) ...("On rationalising the denominator")`
`"RHS" = ((tanθ + secθ + 1)^2)/((tanθ + secθ)^2 - 1)`
`"RHS" = (tan^2θ + sec^2θ + 1 + 2tanθsecθ + 2tanθ + 2secθ)/(tan^2θ + 2tanθ.secθ + sec^2θ - 1) ...{((a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc),((a + b)^2 = a^2 + 2ab + b^2):}`
`"RHS" = ((1 + tan^2θ) + sec^2θ + 2tanθsecθ + 2tanθ + 2secθ)/(tan^2θ + 2tanθ.secθ + (sec^2θ - 1))`
`"RHS" = (sec^2θ + sec^2θ + 2tanθsecθ + 2tanθ + 2secθ)/(tan^2θ + 2tanθ.secθ + tan^2θ)`
`"RHS" = (2sec^2θ + 2tanθsecθ + 2tanθ + 2secθ)/(2tan^2θ + 2tanθ.secθ)`
`"RHS" = [2secθ (tanθ + secθ) + 2(tanθ + secθ)]/[2tanθ(tanθ + secθ)]`
`"RHS" = [(2secθ + 2)(cancel(tanθ + secθ))]/[2tanθ(cancel(tanθ + secθ))]`
`"RHS" = [cancel2(secθ + 1)]/[cancel2(tanθ)]`
`"RHS" = (secθ + 1)/(tanθ)`
`"RHS" = (secθ + 1)/(tanθ) × (secθ - 1)/(secθ - 1)`
`"RHS" = (sec^2θ - 1)/(tanθ(secθ - 1)) ...[(a - b)(a + b) = a^2 - b^2]`
`"RHS" = (tan^cancel2θ)/(canceltanθ(secθ - 1))`
`"RHS" = tanθ/(secθ - 1)`
LHS = `"tanθ"/("secθ" – 1)`
LHS = RHS
Hence proved.
उत्तर २
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
LHS = `"tanθ"/("secθ" – 1)`
`"LHS" = "tanθ"/("secθ" – 1) × ("secθ" + 1)/("secθ"+ 1) ...("On rationalising the denominator")`
`"LHS" = ("tanθ"("secθ" + 1))/("sec"^2θ" - 1) ...[(a + b)(a - b) = a^2 - b^2]`
`"LHS" = ("tanθ"("secθ" + 1))/("tan"^2θ") ...{(∵ 1 + tan^2θ = sec^2θ),(∵ sec^2θ - 1 = tan^2θ):}`
`"LHS" = ("secθ" + 1)/"tanθ"`
∴ `"tanθ"/("secθ" – 1) = ("secθ" + 1)/"tanθ"`
∴ By theorem on equal ratios,
`"tanθ"/("secθ" – 1) = ("secθ" + 1)/"tanθ" = (tanθ + (secθ + 1))/((tanθ) + secθ - 1)`
`"LHS" = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
`"RHS" = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
LHS = RHS
Hence proved.
संबंधित प्रश्न
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Prove that `cosA/(1+sinA) + tan A = secA`
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
if `x/a cos theta + y/b sin theta = 1` and `x/a sin theta - y/b cos theta = 1` prove that `x^2/a^2 + y^2/b^2 = 2`
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
From the figure find the value of sinθ.
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
If tan θ × A = sin θ, then A = ?
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
