Question

verify:-
[tan^2x/(tan^2x-1)]+[cosec^2x/(sec^2x-cosec^2x)]=1/(sin^2x-cos^2x)

Answer 1

Answer on Question #62279 – Math – Trigonometry

Question

Verify:

$$[ \tan^ {2} x / (\tan^ {2} x - 1) ] + [ \csc^ {2} x - \csc^ {2} x ] = 1 / (\sin^ {2} x - \cos^ {2} x)$$

Solution

Let's transform the left-hand side of the expression (1):

$$\begin{array}{l} \frac {\tan^ {2} x}{\tan^ {2} x - 1} + \frac {\csc^ {2} x}{\sec^ {2} x - \csc^ {2} x} = \frac {\sin^ {2} x}{\cos^ {2} x \cdot \left(\frac {\sin^ {2} x}{\cos^ {2} x} - 1\right)} + \frac {\frac {1}{\sin^ {2} x}}{\frac {1}{\cos^ {2} x} - \frac {1}{\sin^ {2} x}} = \\ = \frac {\sin^ {2} x}{\cos^ {2} x \cdot \frac {\sin^ {2} x - \cos^ {2} x}{\cos^ {2} x}} + \frac {\frac {1}{\sin^ {2} x}}{\frac {\sin^ {2} x - \cos^ {2} x}{\cos^ {2} x \cdot \sin^ {2} x}} = \frac {\sin^ {2} x}{\sin^ {2} x - \cos^ {2} x} + \frac {1}{\sin^ {2} x} \cdot \frac {\cos^ {2} x \cdot \sin^ {2} x}{\sin^ {2} x - \cos^ {2} x} = \\ = \frac {\sin^ {2} x}{\sin^ {2} x - \cos^ {2} x} + \frac {\cos^ {2} x}{\sin^ {2} x - \cos^ {2} x} = \frac {\sin^ {2} x + \cos^ {2} x}{\sin^ {2} x - \cos^ {2} x} = \frac {1}{\sin^ {2} x - \cos^ {2} x}. \\ \end{array}$$

We can see that we get the right-hand side of the expression (1). It means that the identity (1) is proved.

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