English

Prove the Following Trigonometric Identities. 1/(Sec a + Tan A) - 1/Cos a = 1/Cos a - 1/(Sec a - Tan A)

Advertisements
Advertisements

Questions

Prove the following trigonometric identities.

`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Prove the following:

`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Theorem
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.

shaalaa.com
  Is there an error in this question or solution?
Chapter 18: Trigonometric identities - Exercise 18A [Page 424]

APPEARS IN

Nootan Mathematics [English] Class 10 ICSE
Chapter 18 Trigonometric identities
Exercise 18A | Q 25. | Page 424
R.D. Sharma Mathematics [English] Class 10
Chapter 11 Trigonometric Identities
Exercise 11.1 | Q 48 | Page 45

RELATED QUESTIONS

`"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`

 


Express the ratios cos A, tan A and sec A in terms of sin A.


Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`

[Hint: Write the expression in terms of sinθ and cosθ]


Prove the following trigonometric identities:

`(1 - cos^2 A) cosec^2 A = 1`


Prove the following trigonometric identities.

sec A (1 − sin A) (sec A + tan A) = 1


Prove the following trigonometric identities.

(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1


Prove the following trigonometric identities

tan2 A + cot2 A = sec2 A cosec2 A − 2


Prove the following identities:

(1 + cot A – cosec A)(1 + tan A + sec A) = 2


Prove the following identities:

`cosA/(1 + sinA) + tanA = secA`


Prove the following identities:

`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`


`cosec theta (1+costheta)(cosectheta - cot theta )=1`


`cot^2 theta - 1/(sin^2 theta ) = -1`a


`sec theta (1- sin theta )( sec theta + tan theta )=1`


`sin theta/((cot theta + cosec  theta)) - sin theta /( (cot theta - cosec  theta)) =2`


Write the value of`(tan^2 theta  - sec^2 theta)/(cot^2 theta - cosec^2 theta)`


If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`


If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.


Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`


Write the value of cosec2 (90° − θ) − tan2 θ. 


Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\] 


What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]


If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2. 


If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ. 


cos4 A − sin4 A is equal to ______.


(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to


(sec A + tan A) (1 − sin A) = ______.


Prove that: 
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1


Simplify 

sin A `[[sinA   -cosA],["cos A"  " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`


Prove the following identity : 

`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`


Prove the following identity : 

`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`


Prove the following identity : 

`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`


Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`


For ΔABC , prove that : 

`sin((A + B)/2) = cos"C/2`


Without using trigonometric identity , show that :

`cos^2 25^circ + cos^2 65^circ = 1`


There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.


If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2


Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.


Prove that:  `1/(sec θ - tan θ) = sec θ + tan θ`.


Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.


If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1


If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4


a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to


sin2θ + sin2(90 – θ) = ?


Prove that `sqrt(sec^2 theta + "cosec"^2 theta) = tan theta + cot theta`


If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.


Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`


If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.


tan θ × `sqrt(1 - sin^2 θ)` is equal to:


If sin A = `1/2`, then the value of sec A is ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×