Advertisements
Advertisements
Question
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Advertisements
Solution
`"LHS" = sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ))`
Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get,
`"LHS" = sqrt((1 + sin θ)/(1 - sin θ) × (1 + sin θ)/(1 + sin θ)) + sqrt((1 - sin θ)/(1 + sin θ) × (1 - sin θ)/(1 - sin θ))`
`"LHS" = sqrt((1 + sin θ)^2/(1 - sin^2 θ)) + sqrt((1 - sin θ)^2/(1 - sin^2 θ))`
`"LHS" = sqrt((1 + sin^2θ)/(1 - sin^2 θ)) + sqrt((1 - sin^2θ)/(1 - sin^2 θ))`
`"LHS" = sqrt((1 + sin^2θ)/(cos^2 θ)) + sqrt((1 - sin^2θ)/(cos^2 θ))`
`"LHS" = (1 + sin θ)/(cos θ) + (1 - sin θ)/(cos θ)`
`"LHS" = (1 + cancel(sin θ) + 1 -cancel(sin θ))/(cos θ)`
LHS = `2/(cos θ)`
LHS = 2. `1/(cos θ)`
LHS = 2. sec θ
RHS = 2. sec θ
LHS = RHS
Hence proved.
APPEARS IN
RELATED QUESTIONS
If sinθ + sin2 θ = 1, prove that cos2 θ + cos4 θ = 1
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`1 - sin^2A/(1 + cosA) = cosA`
Prove the following identities:
`(1 - 2sin^2A)^2/(cos^4A - sin^4A) = 2cos^2A - 1`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`sin theta / ((1+costheta))+((1+costheta))/sin theta=2cosectheta`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
If tan A =` 5/12` , find the value of (sin A+ cos A) sec A.
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Simplify : 2 sin30 + 3 tan45.
Write True' or False' and justify your answer the following:
\[ \cos \theta = \frac{a^2 + b^2}{2ab}\]where a and b are two distinct numbers such that ab > 0.
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
Prove that
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
Prove the following that:
`tan^3θ/(1 + tan^2θ) + cot^3θ/(1 + cot^2θ)` = secθ cosecθ – 2 sinθ cosθ
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
