Question

If $\sec \theta =m$ and $\tan \theta =n$, then $\dfrac{1}{m}[\left( m+n \right)+\dfrac{1}{\left( m+n \right)}]$ is equal to ?

Answer 1

In order to solve the question, we need to put the value of both the trigonometric functions then use the trigonometric identities to solve the equations.
Using identities convert the tangent trigonometric function into secant function then taking common trigonometric functions to get the desired result.

Formula used:
${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$
That is the trigonometric identities of secant and tangent functions.
We also use basic square sum identities of two numbers.

Complete step by step solution:

& \sec \theta =m \\\ & \tan \theta =n \\\ \end{aligned}$$ To find the given equation, we put the value of $$m$$ and $$n$$in the below equation; $$\begin{aligned} & \dfrac{1}{m}[\left( m+n \right)+\dfrac{1}{\left( m+n \right)}] \\\ & =\dfrac{1}{\sec \theta }[(\sec \theta +\tan \theta )]+\dfrac{1}{(\sec \theta +\tan \theta )}] \\\ \end{aligned}$$ Solving the fractional part of the equation, We get: $$=\dfrac{1}{\sec \theta }\times \dfrac{[{{(\sec \theta +\tan \theta )}^{2}}+1]}{(\sec \theta +\tan \theta )}$$ Now, solving the square bracket by the two sum square formula and using the Trigonometric identity $${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$$ We get; $$\begin{aligned} & =\dfrac{1}{\sec \theta }\times \dfrac{[({{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\times \sec \theta \times \tan \theta )+1]}{(\sec \theta +\tan \theta )} \\\ & =\dfrac{1}{\sec \theta }\times \dfrac{[(2{{\sec }^{2}}\theta +2\times \sec \theta \times \tan \theta )]}{(\sec \theta +\tan \theta )} \\\ \end{aligned}$$ Taking out the common term and cancelling out the secant function, we get: $$\begin{aligned} & =\dfrac{1}{\sec \theta }\times 2\times \sec \theta \dfrac{(\sec \theta +\tan \theta )}{(\sec \theta +\tan \theta )} \\\ & =2 \\\ \end{aligned}$$ After cancelling out all the trigonometric function we get; So, The value of $$\dfrac{1}{m}[\left( m+n \right)+\dfrac{1}{\left( m+n \right)}]$$ is $$2$$. **Note:** don’t convert the secant and tangent function directly into sine and cos functions as they would make the equation so long and will take more time. Try to take things common rather than putting values as the maximum term will get cancelled out in the last and will give you the desired result of the given equation. Alternative method: we can also solve this by converting the secant and tangent function into sine and cos function and then take the fraction, at the end it too cancels out the trigonometric term and gives the result.