Question

$(\cos A-\csc A{)}^2+(\sin A-\sec A{)}^2=(1-\sec A.\csc A{)}^2$

• A. True
• B. False

Answer 1

The correct option is A True

Given

${(\cos A-\csc A)}^2+{(\sin A-\sec A)}^2={(1-\sec A\csc A)}^2$

Taking LHS

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

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

$⟹({\sin }^{2}A+{\cos }^{2}A)+{\csc }^{2}A+{\sec }^{2}A-2\left(\frac{\cos A}{\sin A}\right)-2\left(\frac{\sin A}{\cos A}\right)$

$⟹1+\frac{1}{{\sin }^{2}A}+\frac{1}{{\cos }^{2}A}-2\left(\frac{{\cos }^{2}A+{\sin }^2}{\sin A\cos A}\right)$

$⟹1+\left(\frac{{\cos }^{2}A+{\sin }^2}{{\sin }^{2}A{\cos }^{2}A}\right)-2\left(\frac{1}{\sin A\cos A}\right)$

$⟹1+\left(\frac{1}{{\sin }^{2}A{\cos }^{2}A}\right)-2\sec A\csc A$

$⟹1+{\sec }^{2}A{\csc }^{2}A-2\sec A\csc A$

$⟹{(1-\sec A\csc A)}^2$

$\therefore$ LHS $=$ RHS

Hence Proved.