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Question: Prove that \[\left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right) = \cos A\] ?...

Prove that (sec⁡A+tan⁡A)(1−sin⁡A)=cos⁡A\left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right) = \cos A ?

Explanation

Solution

Hint : Here in this question, we have to prove the given trigonometric function by showing the left hand side is equal to the right hand side (i.e., L.H.S=R.H.SL.H.S = R.H.S ). To solve this, we have to consider L.H.S and simplify by using a reciprocal definition of trigonometric ratios and by basic arithmetic operation to get the required solution.

Complete step-by-step answer :
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
Prove that
⇒(sec⁡A+tan⁡A)(1−sin⁡A)=cos⁡A\Rightarrow \left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right) = \cos A --------(1)
Divide both side by (1−sin⁡A)\left( {1 - \sin A} \right) , then
⇒(sec⁡A+tan⁡A)=cos⁡A(1−sin⁡A)\Rightarrow \left( {\sec A + \tan A} \right) = \dfrac{{\cos A}}{{\left( {1 - \sin A} \right)}} --------(2)
Consider Left hand side of equation (1) (L.H.S)
⇒sec⁡A+tan⁡A\Rightarrow \sec A + \tan A --------(3)
Let us by the definition of trigonometric ratios:
Secant is the reciprocal of the cosine ratio i.e., : cos⁡θ=1sec⁡θ\cos \theta = \dfrac{1}{{\sec \theta }}
And tangent is the ratio between the sine and cosine trigonometric ratios i.e., : tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} .
On substituting in equation (3), we have
⇒1cos⁡A+sin⁡Acos⁡A\Rightarrow \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}
Take cos⁡A\cos A as LCM, then
⇒1+sin⁡Acos⁡A\Rightarrow \dfrac{{1 + \sin A}}{{\cos A}}
Multiply and divided by (1−sin⁡A)\left( {1 - \sin A} \right) , then we have
⇒1+sin⁡Acos⁡A×(1−sin⁡A)(1−sin⁡A)\Rightarrow \dfrac{{1 + \sin A}}{{\cos A}} \times \dfrac{{\left( {1 - \sin A} \right)}}{{\left( {1 - \sin A} \right)}}
⇒(1+sin⁡A)(1−sin⁡A)cos⁡A(1−sin⁡A)\Rightarrow \dfrac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)}}{{\cos A\left( {1 - \sin A} \right)}}
Apply an algebraic identity: a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) in numerator.
Here a=1a = 1 and b=sin⁡Ab = \sin A , then we have
⇒12−sin⁡2Acos⁡A(1−sin⁡A)\Rightarrow \dfrac{{{1^2} - {{\sin }^2}A}}{{\cos A\left( {1 - \sin A} \right)}}
⇒1−sin⁡2Acos⁡A(1−sin⁡A)\Rightarrow \dfrac{{1 - {{\sin }^2}A}}{{\cos A\left( {1 - \sin A} \right)}}
By using the trigonometric identity: sin⁡2θ+cos⁡2θ=1⇒cos⁡2θ=1−sin⁡2θ{\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta , then
⇒cos⁡2Acos⁡A(1−sin⁡A)\Rightarrow \dfrac{{{{\cos }^2}A}}{{\cos A\left( {1 - \sin A} \right)}}
On cancelling the like terms in both numerator and denominator, we get
⇒cos⁡A1−sin⁡A\Rightarrow \dfrac{{\cos A}}{{1 - \sin A}}
⇒R.H.S\Rightarrow R.H.S
Therefore, L.H.S=R.H.SL.H.S = R.H.S
⇒sec⁡A+tan⁡A=cos⁡A(1−sin⁡A)\Rightarrow \sec A + \tan A = \dfrac{{\cos A}}{{\left( {1 - \sin A} \right)}}
Hence, proved.

Note : When solving the trigonometry based questions, we have to know the definitions of ratios and always remember the standard angles and formulas are useful for solving certain integration problems where a double formula may make things much simpler to solve. Thus, in math as well as in physics, these formulae are useful to derive many important identities.