Question

Prove that: \[\sqrt{{{\sec }^{2}}A+\text{cose}{{\text{c}}^{2}}A}=\tan A+\cot A\]

Answer 1

Hint: We will begin with the left hand side of the equation (1) and then we will solve it by first converting the terms in sin and cos and then we will take the LCM and finally we will get the answer which will be equal to the right hand side of the expression. We will use the formula \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] to prove the expression.

Complete step-by-step answer:

It is mentioned in the question that \[\sqrt{{{\sec }^{2}}A+\text{cose}{{\text{c}}^{2}}A}=\tan A+\cot A.......(1)\]

Now beginning with the left hand side of the equation (1) we get,

\[\Rightarrow \sqrt{{{\sec }^{2}}A+\text{cose}{{\text{c}}^{2}}A}.....(2)\]

Now converting sec and cosec in equation (2) in terms of cos and sin and hence we get,

\[\Rightarrow \sqrt{\dfrac{1}{{{\cos }^{2}}A}+\dfrac{1}{{{\sin }^{2}}A}}.....(3)\]

Taking the LCM in equation (3) and simplifying we get,

\[\Rightarrow \sqrt{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\cos }^{2}}A{{\sin }^{2}}A}}.....(4)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (4) we get,

\[\begin{align}

  & \Rightarrow \sqrt{\dfrac{1}{{{\cos }^{2}}A{{\sin }^{2}}A}} \\

 & \Rightarrow \dfrac{1}{\cos A\sin A}......(5) \\

\end{align}\]

Now substituting \[{{\sin }^{2}}A+{{\cos }^{2}}A\] in place of 1 in equation (5) we get,

\[\Rightarrow \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\sin A}......(6)\]

Now dividing the numerator terms by the denominator term separately in equation (6) we get,

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{\cos A\sin A}+\dfrac{{{\cos }^{2}}A}{\cos A\sin A}......(7)\]

Now cancelling the similar terms in equation (7) and simplifying we get,

\[\Rightarrow \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}=\tan A+\cot A\]

Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (5) but here the key is to substitute \[{{\sin }^{2}}A+{{\cos }^{2}}A\] in place of 1. Then we need to divide each term in the numerator separately by the denominator in equation (6) to get the left hand side equal to the right hand side of the equation (1).