| The value of \[{{\sec }^{4}}A\,(1-{{\sin }^{4}}A)-2\,\,{{\tan }^{2}}A\] is |
A) \[\frac{1}{2}\]
B) 1
C) 0
D) 2
Correct Answer: B
Solution :
| \[{{\sec }^{4}}A\,(1-{{\sin }^{4}}A)-2{{\tan }^{2}}A\] |
| \[={{\sec }^{4}}A\,[(1-{{\sin }^{2}}A)[(1+{{\sin }^{2}}A)]-2ta{{n}^{2}}A\] |
| \[={{\sec }^{4}}\cdot co{{s}^{2}}A\,\,(1+{{\sin }^{2}}A)-2ta{{n}^{2}}A\] |
| \[[\because 1-{{\sin }^{2}}A={{\cos }^{2}}A]\] |
| \[={{\sec }^{2}}A+\,(1+{{\sin }^{2}}A)-2ta{{n}^{2}}A\] |
| \[={{\sec }^{2}}A+{{\sec }^{2}}A{{\sin }^{2}}A-2{{\tan }^{2}}A\] |
| \[=1+ta{{n}^{2}}A+\frac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-2{{\tan }^{2}}A\] |
| \[=1-{{\tan }^{2}}A+{{\tan }^{2}}A=1\] |
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