Question

If the trigonometric expression as $\sec \left( \alpha \right)+\tan \left( \alpha \right)=m,$ then ${{\left( \sec \left( \alpha \right) \right)}^{4}}-{{\left( \tan \left( \alpha \right) \right)}^{4}}-2\sec \left( \alpha \right)\tan \left( \alpha \right)$ is
(A) ${{m}^{2}}$
(B) $-{{m}^{2}}$
(C) $\dfrac{1}{{{m}^{2}}}$
(D) $-\dfrac{1}{{{m}^{2}}}$

Answer 1

Here to solve this type of problem always make the expression in given equation terms. Apply ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and trigonometric equations to solve this problem. like${{\left[ {{\left( \sec \left( \alpha \right) \right)}^{2}} \right]}^{2}}-{{\left[ {{\left( \tan \left( \alpha \right) \right)}^{2}} \right]}^{2}}$and then we will compare this term with the above formula and then we will solve the above question. Using this formula, we will reduce the given trigonometric term and get to the final answer.

Complete step-by-step solution:
So, here in the above question we have, Given that the $\sec \left( \alpha \right)+\tan \left( \alpha \right)=m$
$\Rightarrow {{\left( \sec \left( \alpha \right) \right)}^{4}}-{{\left( \tan \left( \alpha \right) \right)}^{4}}-2\sec \left( \alpha \right)\tan \left( \alpha \right)$
We can also write the above expression as
$\Rightarrow {{\left[ {{\left( \sec \left( \alpha \right) \right)}^{2}} \right]}^{2}}-{{\left[ {{\left( \tan \left( \alpha \right) \right)}^{2}} \right]}^{2}}-2\sec \left( \alpha \right)\tan \left( \alpha \right)$
Assume $a={{\left( \sec \left( \alpha \right) \right)}^{2}}$and $b={{\left( \tan \left( \alpha \right) \right)}^{2}}$ apply formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ after this we will put the values in the formulas and get the final answer
$\Rightarrow \left[ {{\left( \sec \left( \alpha \right) \right)}^{2}}+{{\left( \tan \left( \alpha \right) \right)}^{2}} \right]\left[ {{\left( \sec \left( \alpha \right) \right)}^{2}}-{{\left( \tan \left( \alpha \right) \right)}^{2}} \right]-2\sec \left( \alpha \right)\tan \left( \alpha \right)$
As we know the trigonometric expression ${{\left( \sec \left( \alpha \right) \right)}^{2}}-{{\left( \tan \left( \alpha \right) \right)}^{2}}=1$ with the help of this formula we will get to the final answer

& \Rightarrow {{\left( \sec \left( \alpha \right) \right)}^{2}}+{{\left( \tan \left( \alpha \right) \right)}^{2}}-2\sec \left( \alpha \right)\tan \left( \alpha \right) \\\ & \Rightarrow {{\left( \sec \left( \alpha \right)-\tan \left( \alpha \right) \right)}^{2}} \\\ \end{aligned}$$ As we have the formula $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ we will reduce the above equation. The trigonometric expression $${{\left( \sec \left( \alpha \right) \right)}^{2}}-{{\left( \tan \left( \alpha \right) \right)}^{2}}=1$$ and assume $$a=\sec \left( \alpha \right)$$ and $$b=\tan \left( \alpha \right)$$ then we will apply this formula $${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$$. $$\begin{aligned} & {{\left( \sec \left( \alpha \right) \right)}^{2}}-{{\left( \tan \left( \alpha \right) \right)}^{2}}=1 \\\ & \left( \sec \left( \alpha \right)-\tan \left( \alpha \right) \right)\left( \sec \left( \alpha \right)+\tan \left( \alpha \right) \right)=1 \\\ & \sec \left( \alpha \right)-\tan \left( \alpha \right)=\dfrac{1}{\left( \sec \left( \alpha \right)+\tan \left( \alpha \right) \right)} \\\ \end{aligned}$$ Also, we have Given that $$\sec \left( \alpha \right)+\tan \left( \alpha \right)=m$$ in the question we will put this in the above so that we will get to the final answer. $$\sec \left( \alpha \right)-\tan \left( \alpha \right)=\dfrac{1}{m}$$ Now we will squaring to the both side then we will get, $$\begin{aligned} & \Rightarrow {{\left( \sec \left( \alpha \right)-\tan \left( \alpha \right) \right)}^{2}}={{\left( \dfrac{1}{m} \right)}^{2}} \\\ & \Rightarrow \dfrac{1}{{{m}^{2}}} \\\ \end{aligned}$$ **Hence, the Option (C) is the correct option.** **Note:** We can solve this question by various methods and we will get the same answer for every method. the Other method is to change everything in sin and cosine form like we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$ with the help of this substitution we can solve this question and then find the values of either sin or cosine. We will get our answer.