Question

$(\sin A+\csc A{)}^2+(\cos A+\sec A{)}^2=7+{\tan }^{2}A+{\cot }^{2}A$.

• A. True
• B. False

Answer 1

The correct option is A True

${(\sin A+\csc A)}^2+{(\cos A+\sec A)}^2$

$={\sin }^{2}A+{\csc }^{2}A+2\sin A\csc A+{\cos }^{2}A+{\sec }^{2}A+2\cos A\sec A$

$={\sin }^{2}A+{\cos }^{2}A+{\csc }^{2}A+2\times 1+{\sec }^{2}A+2\times 1$ since $\sin A\csc A=1$ and $\cos A\sec A=1$

$=1+{\csc }^{2}A+2+{\sec }^{2}A+2$ since ${\sin }^{2}A+{\cos }^{2}A=1$

$=5+1+{\cot }^{2}A+1+{\tan }^{2}A$ since ${\csc }^{2}A=1+{\cot }^{2}A$ and ${\sec }^{2}A=1+{\tan }^{2}A$

$=7+{\tan }^{2}A+{\cot }^{2}A$