Question 141

What is the value of $$\frac{(\tan^{2}x-\sin^{2}x)}{\sec^{2}x}$$ ?

Solution

Expression : $$\frac{(\tan^{2}x-\sin^{2}x)}{\sec^{2}x}$$

= $$\frac{(\frac{sin^{2}x}{cos^2x}-sin^{2}x)}{sec^{2}x}$$

= $$\frac{sin^2x}{sec^2x}(\frac{1}{cos^2x}-1)$$

= $$sin^2x cos^2x(\frac{1-cos^2x}{cos^2x})$$

= $$sin^2x\times(sin^2x)=sin^4x$$

=> Ans - (A)

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