Question

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

Answer 1

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., \[L.H.S = R.H.S\] ). 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
 \[ \Rightarrow \left( {\sec A + \tan A} \right)\left( {1 - \sin A} \right) = \cos A\] --------(1)
Divide both side by \[\left( {1 - \sin A} \right)\] , then
 \[ \Rightarrow \left( {\sec A + \tan A} \right) = \dfrac{{\cos A}}{{\left( {1 - \sin A} \right)}}\] --------(2)
Consider Left hand side of equation (1) (L.H.S)
 \[ \Rightarrow \sec A + \tan A\] --------(3)
Let us by the definition of trigonometric ratios:
Secant is the reciprocal of the cosine ratio i.e., : \[\cos \theta = \dfrac{1}{{\sec \theta }}\]
And tangent is the ratio between the sine and cosine trigonometric ratios i.e., : \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] .
On substituting in equation (3), we have
 \[ \Rightarrow \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}\]
Take \[\cos A\] as LCM, then
 \[ \Rightarrow \dfrac{{1 + \sin A}}{{\cos A}}\]
Multiply and divided by \[\left( {1 - \sin A} \right)\] , then we have
 \[ \Rightarrow \dfrac{{1 + \sin A}}{{\cos A}} \times \dfrac{{\left( {1 - \sin A} \right)}}{{\left( {1 - \sin A} \right)}}\]
 \[ \Rightarrow \dfrac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)}}{{\cos A\left( {1 - \sin A} \right)}}\]
Apply an algebraic identity: \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in numerator.
Here \[a = 1\] and \[b = \sin A\] , then we have
 \[ \Rightarrow \dfrac{{{1^2} - {{\sin }^2}A}}{{\cos A\left( {1 - \sin A} \right)}}\]
 \[ \Rightarrow \dfrac{{1 - {{\sin }^2}A}}{{\cos A\left( {1 - \sin A} \right)}}\]
By using the trigonometric identity: \[{\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta \] , then
 \[ \Rightarrow \dfrac{{{{\cos }^2}A}}{{\cos A\left( {1 - \sin A} \right)}}\]
On cancelling the like terms in both numerator and denominator, we get
 \[ \Rightarrow \dfrac{{\cos A}}{{1 - \sin A}}\]
 \[ \Rightarrow R.H.S\]
Therefore, \[L.H.S = R.H.S\]
 \[ \Rightarrow \sec A + \tan A = \dfrac{{\cos A}}{{\left( {1 - \sin A} \right)}}\]
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.