Solveeit Logo
Login

Question

Question: To prove that \(\dfrac{\tan x}{1-\cot x}+\dfrac{\cot x}{1-\tan x}=1+\sec x\cos ecx\)....

To prove that tan⁡x1−cot⁡x+cot⁡x1−tan⁡x=1+sec⁡xcos⁡ecx\dfrac{\tan x}{1-\cot x}+\dfrac{\cot x}{1-\tan x}=1+\sec x\cos ecx.

Explanation

Solution

Hint: To solve this problem, we express the value of tanx and cotx in terms of sinx and cosx. We use the trigonometric property that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} and cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}. We would then substitute these values in the above expression to solve the LHS expression.

Complete step-by-step solution -
We have to solve the expression tan⁡x1−cot⁡x+cot⁡x1−tan⁡x\dfrac{\tan x}{1-\cot x}+\dfrac{\cot x}{1-\tan x} and prove its equality to the expression 1+sec⁡xcos⁡ecx1+\sec x\cos ecx. We now express the value of tanx and cotx in terms of sinx and cosx, thus, we get,
=sin⁡xcos⁡x1−cos⁡xsin⁡x+cos⁡xsin⁡x1−sin⁡xcos⁡x=\dfrac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}}+\dfrac{\dfrac{\cos x}{\sin x}}{1-\dfrac{\sin x}{\cos x}}
=sin⁡xcos⁡xsin⁡x−cos⁡xsin⁡x+cos⁡xsin⁡xcos⁡x−sin⁡xcos⁡x=\dfrac{\dfrac{\sin x}{\cos x}}{\dfrac{\sin x-\cos x}{\sin x}}+\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x-\sin x}{\cos x}}
Re-arranging, we get,
=sin⁡2xcos⁡x(sin⁡x−cos⁡x)+cos⁡2xsin⁡x(cos⁡x−sin⁡x)=\dfrac{{{\sin }^{2}}x}{\cos x(\sin x-\cos x)}+\dfrac{{{\cos }^{2}}x}{\sin x(\cos x-\sin x)}
Further, taking 1(sin⁡x−cos⁡x)\dfrac{1}{(\sin x-\cos x)} as the common term, we get,
=1(sin⁡x−cos⁡x)(sin⁡2xcos⁡x−cos⁡2xsin⁡x)=\dfrac{1}{(\sin x-\cos x)}\left( \dfrac{{{\sin }^{2}}x}{\cos x}-\dfrac{{{\cos }^{2}}x}{\sin x} \right)
=1(sin⁡x−cos⁡x)(sin⁡3x−cos⁡3xsin⁡xcos⁡x)=\dfrac{1}{(\sin x-\cos x)}\left( \dfrac{{{\sin }^{3}}x-{{\cos }^{3}}x}{\sin x\cos x} \right)
Now, we use the algebraic identity that a3−b3=(a−b)(a2+ab+b2){{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}}). In this case, a = sinx and b = cosx. Thus, we have,
=1(sin⁡x−cos⁡x)((sin⁡x−cos⁡x)(sin⁡2x+sin⁡xcos⁡x+cos⁡2x)sin⁡xcos⁡x)=\dfrac{1}{(\sin x-\cos x)}\left( \dfrac{\left( \sin x-\cos x \right)\left( {{\sin }^{2}}x+\sin x\cos x+{{\cos }^{2}}x \right)}{\sin x\cos x} \right)
We can now cancel the term (sinx – cosx) from both the numerator and the denominator. Thus, we have,
=((sin⁡2x+sin⁡xcos⁡x+cos⁡2x)sin⁡xcos⁡x)=\left( \dfrac{\left( {{\sin }^{2}}x+\sin x\cos x+{{\cos }^{2}}x \right)}{\sin x\cos x} \right)
Now dividing each term on the numerator by sinxcosx, we have,
= sin⁡2xsin⁡xcos⁡x+sin⁡xcos⁡xsin⁡xcos⁡x+cos⁡2xsin⁡xcos⁡x\dfrac{{{\sin }^{2}}x}{\sin x\cos x}+\dfrac{\sin x\cos x}{\sin x\cos x}+\dfrac{{{\cos }^{2}}x}{\sin x\cos x}
=sin⁡xcos⁡x+1+cos⁡xsin⁡x=\dfrac{\sin x}{\cos x}+1+\dfrac{\cos x}{\sin x}
=1+sin⁡2x+cos⁡2xsin⁡xcos⁡x=1+\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x}
Now, using the trigonometric identity that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, we have,
=1+1sin⁡xcos⁡x=1+\dfrac{1}{\sin x\cos x}
= 1 + (cos ecx)(secx) -- (A)
Since, cos⁡ecx=1sin⁡x\cos ecx=\dfrac{1}{\sin x} and secx=1cos⁡xsecx=\dfrac{1}{\cos x}.
Since, the expression (A) is the same as the RHS expression. Thus, LHS = RHS. Hence, proved.

Note: Another alternative way to solve this problem is by solving the RHS. In this case, we backtrack the RHS expression by following the above solution shown in the opposite direction to arrive at the LHS term. However, it is an extremely difficult task to intuitively arrive at the LHS expression by starting from the RHS expression.