Question

Prove that \[\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}} = 2\sec A\]?

Answer 1

Hint: Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratio. Hence by using the simple calculations we are going to prove the given inequality.

Complete step by step solution:
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
Now consider the given inequality \[\dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}} = 2\sec A\]
Now we consider the LHS
\[ \Rightarrow \dfrac{{\cos A}}{{1 + \sin A}} + \dfrac{{1 + \sin A}}{{\cos A}}\]
No we take the LCM for the both terms and this is written as
\[ \Rightarrow \dfrac{{\cos A.\cos A + (1 + \sin A).(1 + \sin A)}}{{(1 + \sin A)\cos A}}\]
On simplifying we have
\[ \Rightarrow \dfrac{{{{\cos }^2}A + {{(1 + \sin A)}^2}}}{{\cos A(1 + \sin A)}}\]
The above inequality, in the numerator it is in the form of \[{(a + b)^2}\], we have standard algebraic identity for this and it is given as \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] Using this identity the above inequality is written as
\[ \Rightarrow \dfrac{{{{\cos }^2}A + 1 + {{\sin }^2}A + 2\sin A}}{{\cos A(1 + \sin A)}}\]
From the trigonometry identities we have \[{\sin ^2}A + {\cos ^2}A = 1\]
Therefore we have
\[ \Rightarrow \dfrac{{1 + 1 + 2\sin A}}{{\cos A(1 + \sin A)}}\]
On simplifying we have
\[ \Rightarrow \dfrac{{2 + 2\sin A}}{{\cos A(1 + \sin A)}}\]
Take 2 as a common we have
\[ \Rightarrow \dfrac{{2(1 + \sin A)}}{{\cos A(1 + \sin A)}}\]
The term \[(1 + \sin A)\] gets cancelled since it is present on both the numerator and denominator.
\[ \Rightarrow \dfrac{2}{{\cos A}}\]
The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosecant is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent. From the reciprocal of the trigonometry ratios. Now the inequality is written as
\[ \Rightarrow 2\sec A\]
\[ \Rightarrow RHS\]
Hence proved.

Note:
The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.