Advertisements
Advertisements
Question
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Advertisements
Solution
In the given question, we need to prove `1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Here, we will first solve the L.H.S.
Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get
`1/(sec A + tan A) - 1/cos A = 1/(1/cos A + sin A/cos A) - (1/cos A)`
`= 1/(((1 + sin A)/cos A)) - (1/cos A)`
`= (cos A/(1 + sin A)) - (1/cos A)`
`= (cos^2 A - (1 + sin A))/((1 + sin A)(cos A))`
On further solving, we get
`(cos^2 A -(1 + sin A))/((1 + sin A)(cos A)) = (cos^2 A - 1 - sin A)/((1 + sin A)(cos A))`
`= (-sin^2 A - sin A)/((1 + sin A)(cos A))` (Using `sin^2 theta = 1 - cos^2 theta)`
`= (-sin A(sin A + 1))/((1 + sin A)(cos A))`
`= (-sin A)/cos A`
= -tan A
Similarly we solve the R.H.S.
`((1 - sin A) - cos^2 A)/((cos A)(1 - sin^2 A)) = (1 - sin A - cos^2 A)/((cos A)(1 - sin A))`
`= (sin^2 A - sin A)/((cos A)(1 - sin A))` (Using `sin^2 theta = 1- cos^2 theta`)
`= (-sin A(1 - sin A))/((cos A)(1 - sin A))`
`= (-sin A)/cos A`
= - tan A
So, L.H.S = R.H.S
Hence proved.
RELATED QUESTIONS
If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
If sin A + cos A = p and sec A + cosec A = q, then prove that : q(p2 – 1) = 2p.
If x = a cos θ and y = b cot θ, show that:
`a^2/x^2 - b^2/y^2 = 1`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
`(1-cos^2theta) sec^2 theta = tan^2 theta`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
If `sec theta + tan theta = p,` prove that
(i)`sec theta = 1/2 ( p+1/p) (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
If `cos B = 3/5 and (A + B) =- 90° ,`find the value of sin A.
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
\[\frac{x^2 - 1}{2x}\] is equal to
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then\[\frac{x^2}{a^2} + \frac{y^2}{b^2}\]
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
prove that `1/(1 + cos(90^circ - A)) + 1/(1 - cos(90^circ - A)) = 2cosec^2(90^circ - A)`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
Choose the correct alternative:
1 + cot2θ = ?
Prove that `"cosec" θ xx sqrt(1 - cos^2theta)` = 1
Prove that cot2θ × sec2θ = cot2θ + 1
If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
