Question

State whether the following statement is true or false.

$\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

• A. True
• B. False

Answer 1

The correct option is A True

Consider $RHS=\frac{1}{secA-tanA}-\frac{1}{cosA}$

Multiplying by $secA+tanA$ in the numerator and denominator of

first term,

we get $\frac{secA+tanA}{(secA+tanA)(secA-tanA)}-\frac{1}{cosA}$

$=secA+tanA-secA........................(Since,sec²A-tan²A=1)$

$=tanA$

Adding and subtracting $secA$, we get

$=secA+tanA-secA$

$=\frac{1}{cosA}-(secA-tanA)$

Now multiplying and dividing $(secA-tanA)$ by $(secA+tanA)$,

we get

$=\frac{1}{cosA}-\frac{(sec²A-tan²A)}{(secA+tanA)}$

$=\frac{1}{cosA}-\frac{1}{secA+tanA}$

= L.H.S