Advertisements
Advertisements
प्रश्न
`1/((1+ sin θ)) + 1/((1 - sin θ)) = 2 sec^2 θ`
Prove that: `1/(1+ sin θ) + 1/(1 - sin θ) = 2 sec^2 θ`
Advertisements
उत्तर
LHS =`1/((1+ sin θ)) + 1/((1 - sin θ))`
= `((1 - sin θ) + (1 + sin θ))/((1 + sin θ)(1 - sin θ))`
= `2/(1 - sin^2 θ)`
= `2/(cos^2 θ)`
= 2 sec2 θ
= RHS
Hence Proved.
Notes
Students should refer to the answer according to their questions.
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
If tanθ `= 3/4` then find the value of secθ.
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
What is the value of 9cot2 θ − 9cosec2 θ?
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`1/(tanA + cotA) = sinAcosA`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
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
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
