Prove that cot A – cos A cosec A – 1 cot A + cos A cosec A + 1
Step-by-Step Solution
Key Concept: Use the fundamental identities \(\cot A=\dfrac{\cos A}{\sin A}\) and \(\csc A=\dfrac{1}{\sin A}\). Simplify each fraction by bringing them to a common denominator and factor \((1\pm\sin A)\) to cancel.
1. Write the trigonometric functions in terms of \(\sin A\) and \(\cos A\):
$$\cot A = \frac{\cos A}{\sin A},\qquad \csc A = \frac{1}{\sin A}.$$
2. Left–hand side (LHS)
\[
\frac{\cot A-\cos A}{\csc A-1}=\frac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{1}{\sin A}-1}
=\frac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{1-\sin A}{\sin A}}
\]
Multiply numerator and denominator by \(\sin A\) (or simply divide the two fractions):
\[
=\frac{\cos A(1-\sin A)}{\sin A}\times\frac{\sin A}{1-\sin A}=\cos A.
\]
3. Right–hand side (RHS)
\[
\frac{\cot A+\cos A}{\csc A+1}=\frac{\dfrac{\cos A}{\sin A}+\cos A}{\dfrac{1}{\sin A}+1}
=\frac{\dfrac{\cos A+\cos A\sin A}{\sin A}}{\dfrac{1+\sin A}{\sin A}}
\]
Again cancel the common factor \((1+\sin A)\):
\[
=\frac{\cos A(1+\sin A)}{\sin A}\times\frac{\sin A}{1+\sin A}=\cos A.
\]
4. Since both LHS and RHS simplify to the same expression \(\cos A\), the given identity is proved:
\[
\frac{\cot A-\cos A}{\csc A-1}=\frac{\cot A+\cos A}{\csc A+1}=\cos A.
\]
Hence the statement is true.
Correct Answer: Both sides simplify to \(\cos A\); therefore the identity holds.