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Question: Show that \((\sin \theta + \cos \theta )(\tan \theta + \cot \theta ) = \sec \theta + \cos ec\theta \...

Show that (sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+cos⁡ecθ(\sin \theta + \cos \theta )(\tan \theta + \cot \theta ) = \sec \theta + \cos ec\theta

Explanation

Solution

According to given in the question we have to show that (sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+cos⁡ecθ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta ) = \sec \theta + \cos ec\theta so, we will solve the left hand side term
of the given expression which is (sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)(\sin \theta + \cos \theta )(\tan \theta + \cot \theta )
To solve L.H.S. first of all we will try to make the term tan⁡θ\tan \theta in form of sin⁡θ\sin \theta and cos⁡θ\cos \theta with the help of formula as given below:

Formula used: tan⁡θ=sin⁡θcos⁡θ.................(1)\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.................(1)
Now, same as we will try to make the term cot⁡θ\cot \theta in form of sin⁡θ\sin \theta and cos⁡θ\cos \theta with the help of formula as given below:
cot⁡θ=cos⁡θsin⁡θ.................(2)\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}.................(2)
After obtaining the expression in form of sin⁡θ\sin \theta and cos⁡θ\cos \theta we will take the L.C.M. to solve it and after L.C.M. we will obtain the terms sin⁡θ\sin \theta and cos⁡θ\cos \theta in their square form which can be solved with the help of the formula as given below:
sin⁡2θ+cos⁡2θ=1...............(3){\sin ^2}\theta + {\cos ^2}\theta = 1...............(3)

Complete step-by-step answer:
Step 1: First of all we have to make the term tan⁡θ\tan \theta as given in the L.H.S. of the expression in the form of sin⁡θ\sin \theta and cos⁡θ\cos \theta with the help of formula (1) as mentioned in the solution
hint.
=(sin⁡θ+cos⁡θ)(sin⁡θcos⁡θ+cot⁡θ)= (\sin \theta + \cos \theta )\left( {\dfrac{{\sin \theta }}{{\cos \theta }} + \cot \theta } \right)
Step 2: Same as, to make the term cot⁡θ\cot \theta as given in the L.H.S. of the expression in form of sin⁡θ\sin \theta and cos⁡θ\cos \theta with the help of formula (2) as mentioned in the solution hint.
=(sin⁡θ+cos⁡θ)(sin⁡θcos⁡θ+cos⁡θsin⁡θ)= (\sin \theta + \cos \theta )\left( {\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\sin\theta }}} \right)
Step 3: Now, we have to solve the trigonometric expression as obtained in step 2.
=(sin⁡θ+cos⁡θ)(sin⁡2θ+cos⁡2θsin⁡θcos⁡θ)= (\sin \theta + \cos \theta )\left( {\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta \cos\theta }}} \right)
Step 4: Now, to solve the expression as obtained in step 3 we have to use the formula (3) as mentioned in the solution hint.
=(sin⁡θ+cos⁡θ)(1sin⁡θcos⁡θ)= (\sin \theta + \cos \theta )\left( {\dfrac{1}{{\sin \theta \cos \theta }}} \right)
Step 5: Now, we have to multiply and divide the terms of the expression as obtained in the step 4.
=sin⁡θ+cos⁡θsin⁡θcos⁡θ =sin⁡θsin⁡θcos⁡θ+cos⁡θsin⁡θcos⁡θ  = \dfrac{{\sin \theta + \cos \theta }}{{\sin \theta \cos \theta }} \\\ = \dfrac{{\sin \theta }}{{\sin \theta \cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta \cos \theta }} \\\
On solving the obtained expression,
=1cos⁡θ+1sin⁡θ= \dfrac{1}{{\cos \theta }} + \dfrac{1}{{\sin \theta }}…………………..(4)
Step 6: As we know that, sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }} and cos⁡ecθ=1sin⁡θ\cos ec\theta = \dfrac{1}{{\sin \theta }} hence, substituting in the expression (4) as obtained in the step 5.
=sec⁡θ+cos⁡ecθ= \sec \theta + \cos ec\theta

Hence, with the help of the formula (1), (2), and (3) we have proved that (sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+cos⁡ecθ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta ) = \sec \theta + \cos ec\theta

Note: We can also solve the given expression by multiplying each term of the given expression but it will lead us to lots of difficult calculations.
To make the calculations easy we have to use the formula sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 to eliminate the terms like sin⁡2θ{\sin ^2}\theta and cos⁡2θ{\cos ^2}\theta .