Question
$\left[\frac{1-\tan A}{1-\cot A}\right]^2=\tan ^2 A ; \angle A$ is acute
✓
Answer
To prove $\left[\frac{1-\tan A}{1-\cot A}\right]^2=\tan ^2 A$,
Using the left hand side of the equation,
$
\left[\frac{1-\tan A}{1-\cot A}\right]^2
$
Since, $\tan A=\frac{\sin A}{\cos A}$
$\Rightarrow\left[\frac{1-\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}\right]^2$
$\Rightarrow\left[\frac{\frac{\cos A-\sin A}{\cos A}}{\frac{\sin A-\cos A }{\sin A }}\right]^2$
$\Rightarrow\left[\frac{\cos A -\sin A }{\sin A -\cos A } \cdot \frac{\sin A }{\cos A }\right]^2 \Rightarrow\left[-\frac{\sin A }{\cos A }\right]^2=\tan ^2 A$
Since LHS = RHS,
Hence proved, $\left[\frac{1-\tan A }{1-\cot A }\right]^2=\tan ^2 A$
Using the left hand side of the equation,
$
\left[\frac{1-\tan A}{1-\cot A}\right]^2
$
Since, $\tan A=\frac{\sin A}{\cos A}$
$\Rightarrow\left[\frac{1-\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}\right]^2$
$\Rightarrow\left[\frac{\frac{\cos A-\sin A}{\cos A}}{\frac{\sin A-\cos A }{\sin A }}\right]^2$
$\Rightarrow\left[\frac{\cos A -\sin A }{\sin A -\cos A } \cdot \frac{\sin A }{\cos A }\right]^2 \Rightarrow\left[-\frac{\sin A }{\cos A }\right]^2=\tan ^2 A$
Since LHS = RHS,
Hence proved, $\left[\frac{1-\tan A }{1-\cot A }\right]^2=\tan ^2 A$
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