Consider the following equations |
I. \[\text{cose}{{\text{c}}^{2}}x+{{\sec }^{2}}x=\text{cose}{{\text{c}}^{2}}x{{\sec }^{2}}x\] |
II. \[{{\sec }^{2}}x+{{\tan }^{2}}x={{\sec }^{2}}x{{\tan }^{2}}x\] |
III. \[\text{cose}{{\text{c}}^{2}}x+{{\tan }^{2}}x={{\cot }^{2}}x+{{\sec }^{2}}x\] |
Which of the above statements are correct? |
A) Both I and II
B) Both II and III
C) Both I and III
D) I, and III
Correct Answer: C
Solution :
LHS \[=\text{cose}{{\text{c}}^{2}}x+{{\sec }^{2}}x=\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}\] |
\[=\text{cose}{{\text{c}}^{2}}x{{\sec }^{2}}x=RHS\] |
II. LHS\[={{\sec }^{2}}x+{{\tan }^{2}}x=\frac{1+{{\sin }^{2}}x}{{{\cos }^{2}}x}\ne RHS\] |
III. LHS \[=\text{cose}{{\text{c}}^{2}}x+{{\tan }^{2}}x={{\cot }^{2}}x+1+{{\tan }^{2}}x\] |
\[={{\cot }^{2}}x+{{\sec }^{2}}x=RHS\] |
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