Question

To prove that $\dfrac{\tan x}{1-\cot x}+\dfrac{\cot x}{1-\tan x}=1+\sec x\cos ecx$.

Answer 1

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 $\tan x=\dfrac{\sin x}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$. 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 $\dfrac{\tan x}{1-\cot x}+\dfrac{\cot x}{1-\tan x}$ and prove its equality to the expression $1+\sec x\cos ecx$. We now express the value of tanx and cotx in terms of sinx and cosx, thus, we get,
$=\dfrac{\dfrac{\sin x}{\cos x}}{1-\dfrac{\cos x}{\sin x}}+\dfrac{\dfrac{\cos x}{\sin x}}{1-\dfrac{\sin x}{\cos x}}$
$=\dfrac{\dfrac{\sin x}{\cos x}}{\dfrac{\sin x-\cos x}{\sin x}}+\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x-\sin x}{\cos x}}$
Re-arranging, we get,
$=\dfrac{{{\sin }^{2}}x}{\cos x(\sin x-\cos x)}+\dfrac{{{\cos }^{2}}x}{\sin x(\cos x-\sin x)}$
Further, taking \[\dfrac{1}{(\sin x-\cos x)}\] as the common term, we get,
\[=\dfrac{1}{(\sin x-\cos x)}\left( \dfrac{{{\sin }^{2}}x}{\cos x}-\dfrac{{{\cos }^{2}}x}{\sin x} \right)\]
\[=\dfrac{1}{(\sin x-\cos x)}\left( \dfrac{{{\sin }^{3}}x-{{\cos }^{3}}x}{\sin x\cos x} \right)\]
Now, we use the algebraic identity that ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$. In this case, a = sinx and b = cosx. Thus, we have,
\[=\dfrac{1}{(\sin x-\cos x)}\left( \dfrac{\left( \sin x-\cos x \right)\left( {{\sin }^{2}}x+\sin x\cos x+{{\cos }^{2}}x \right)}{\sin x\cos x} \right)\]
We can now cancel the term (sinx – cosx) from both the numerator and the denominator. Thus, we have,
\[=\left( \dfrac{\left( {{\sin }^{2}}x+\sin x\cos x+{{\cos }^{2}}x \right)}{\sin x\cos x} \right)\]
Now dividing each term on the numerator by sinxcosx, we have,
= $\dfrac{{{\sin }^{2}}x}{\sin x\cos x}+\dfrac{\sin x\cos x}{\sin x\cos x}+\dfrac{{{\cos }^{2}}x}{\sin x\cos x}$
$=\dfrac{\sin x}{\cos x}+1+\dfrac{\cos x}{\sin x}$
$=1+\dfrac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x}$
Now, using the trigonometric identity that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we have,
$=1+\dfrac{1}{\sin x\cos x}$
= 1 + (cos ecx)(secx) -- (A)
Since, $\cos ecx=\dfrac{1}{\sin x}$ and $secx=\dfrac{1}{\cos x}$.
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.