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Question: Prove that \( \dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}} \) ....

Prove that 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}} .

Explanation

Solution

Hint : As we know that the above question is related to the trigonometry. Sine, cosine, secant, these are all trigonometric ratios. We will solve the above given question with the help of basic trigonometric formulas. We know that sec⁡θ\sec \theta can also be written in the form of 1cos⁡θ\dfrac{1}{{\cos \theta }} . It is the reciprocal formula of trigonometric ratios.

Complete step-by-step answer :
As per the given question we have: 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}} . We have to prove that the left hand side expression is equal to the right hand expression by using the trigonometric formulas.
We will first take the left hand side of the equation and solve it by putting the value of sec in terms of cosine i.e.
1+sec⁡Asec⁡A=1+1cos⁡A1cos⁡A\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{1 + \dfrac{1}{{\cos A}}}}{{\dfrac{1}{{\cos A}}}} .
By taking the L.C.M we can write it as cos⁡A+1cos⁡A1cos⁡A\dfrac{{\dfrac{{\cos A + 1}}{{\cos A}}}}{{\dfrac{1}{{\cos A}}}} .
By converting it into the simple fraction we have: cos⁡A+1cos⁡A×cos⁡A1=cos⁡A+1\dfrac{{\cos A + 1}}{{\cos A}} \times \dfrac{{\cos A}}{1} = \cos A + 1 . By rationalizing the value: 1+cos⁡A×1−cos⁡A1+cos⁡A1 + \cos A \times \dfrac{{1 - \cos A}}{{1 + \cos A}} .
On further solving (1)2−(cos⁡A)21−cos⁡A=1−cos⁡2A1−cos⁡A\dfrac{{{{(1)}^2} - {{(\cos A)}^2}}}{{1 - \cos A}} = \dfrac{{1 - {{\cos }^2}A}}{{1 - \cos A}} .
We know the trigonometric identity which states that 1−cos⁡2θ=sin⁡2θ1 - {\cos ^2}\theta = {\sin^2} \theta . By applying the same in the above expression we can write it as sin⁡2A1−cos⁡A\dfrac{{{{\sin }^2}A}}{{1 - \cos A}} . This value is equal to the right hand side of the expression.
Hence it is proved that 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{{1 + \sec A}}{{\sec A}} = \dfrac{{{{\sin }^2}A}}{{1 - \cos A}} .

Note : We should note that the difference formula is used in the above i.e. (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} . Trigonometric functions are also called circular functions and these basic functions are also known as trigonometric ratios. There are multiple trigonometric formulas and identities which represent the relation between the functions and enable to find the value of unknown angles.