Question

Prove that: \[\sec \left( \dfrac{3\pi }{2}-\theta \right)\sec \left( \theta -\dfrac{5\pi }{2} \right)+\tan \left( \dfrac{5\pi }{2}+\theta \right)\tan \left( \theta -\dfrac{3\pi }{2} \right)=-1\]

Answer 1

Hint: We will first simplify the terms in the left hand side of the expression in the formula form and then with the help of cofunction identities and periodic identities we will solve this question.

Complete step-by-step answer:

\[\sec \left( \dfrac{3\pi }{2}-\theta \right)\sec \left( \theta -\dfrac{5\pi }{2} \right)+\tan \left( \dfrac{5\pi }{2}+\theta \right)\tan \left( \theta -\dfrac{3\pi }{2} \right)=-1.......(1)\]

We will start with simplifying the left hand side of the equation (1). So,

\[\Rightarrow \sec \left( \dfrac{3\pi }{2}-\theta \right)\sec \left( -\left( \dfrac{5\pi }{2}-\theta \right) \right)+\tan \left( \dfrac{5\pi }{2}+\theta \right)\tan \left( -\left( \dfrac{3\pi }{2}-\theta \right) \right).......(2)\]

Now we know from the trigonometry identities that \[\sec (-\theta )=\sec \theta \] and \[\tan (-\theta )=-\tan \theta \]. So using this information in equation (2) we get,

\[\Rightarrow \sec \left( \dfrac{3\pi }{2}-\theta \right)\sec \left( \dfrac{5\pi }{2}-\theta \right)-\tan \left( \dfrac{5\pi }{2}+\theta \right)\tan \left( \dfrac{3\pi }{2}-\theta \right).......(3)\]

Rearranging the terms in equation (3) and writing \[\dfrac{5\pi }{2}\] as sum of \[2\pi \] and \[\dfrac{\pi }{2}\] we get,

\[\Rightarrow \sec \left( \dfrac{3\pi }{2}-\theta \right)\sec \left( 2\pi +\left( \dfrac{\pi }{2}-\theta \right) \right)-\tan \left( 2\pi +\left( \dfrac{\pi }{2}+\theta \right) \right)\tan \left( \dfrac{3\pi }{2}-\theta \right).......(4)\]

Now we know from the cofunction identities that \[\sec \left( \dfrac{3\pi }{2}-\theta \right)=-\text{cosec}\theta \] and \[\tan \left( \dfrac{3\pi }{2}-\theta \right)=\cot \theta \]. So using this information in equation (4) we get,

\[\Rightarrow -\text{cosec}\theta \sec \left( 2\pi +\left( \dfrac{\pi }{2}-\theta \right) \right)-\tan \left( 2\pi +\left( \dfrac{\pi }{2}+\theta \right) \right)\cot \theta .......(5)\]

We also know from the periodic identities that \[\sec \left( 2\pi +\left( \dfrac{\pi }{2}-\theta \right) \right)=\sec \left( \dfrac{\pi }{2}-\theta \right)\] and \[\tan \left( 2\pi +\left( \dfrac{\pi }{2}+\theta \right) \right)=\tan \left( \dfrac{\pi }{2}+\theta \right)\]. So using this information in equation (5) we get,

\[\Rightarrow -\text{cosec}\theta \sec \left( \dfrac{\pi }{2}-\theta \right)-\tan \left( \dfrac{\pi }{2}+\theta \right)\cot \theta .......(6)\]

Also we know that \[\sec \left( \dfrac{\pi }{2}-\theta \right)=\text{cosec}\theta \] and \[\tan \left( \dfrac{\pi }{2}+\theta \right)=-\cot \theta \]. So using this information in equation (6)

we get,

\[\Rightarrow -\text{cosec}\theta \text{cosec }\!\!\theta\!\!\text{ +cot}\theta \cot \theta .......(7)\]

Now squaring the terms in equation (7) we get,

\[\Rightarrow -\text{cose}{{\text{c}}^{2}}\theta \text{+co}{{\text{t}}^{2}}\theta .......(8)\]

Now we know that \[\text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1\]. So using this information in equation (8) we get,

\[\Rightarrow -\text{cose}{{\text{c}}^{2}}\theta \text{+co}{{\text{t}}^{2}}\theta =-1\]

And hence this is equal to the right hand side of the equation (1). Thus we have proved the equation (1).

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identities and the periodic identities as we can write cosec in place of –cosec and cot in place of –cot. Also if we don’t remember the formula then we can get confused about how to proceed further after equation (2).