Question

Which of the following options is equal to $\frac{\text{tan A}}{1-\text{cot A}}+\frac{\text{cot A}}{1-\text{tan A}}$ ?

• A. $1+\text{cosec A sec A}$
• B. $1+\text{tan A}+\text{cot A}$
• C. $\text{cosec A sec A}$
• D. $\text{tan A + cot A}$

Answer 1

Answer: (A), (B)

The correct options are
A

$1+\text{cosec A sec A}$

B

$1+\text{tan A}+\text{cot A}$

$\frac{\text{tan A}}{1-\text{cot A}}+\frac{\text{cot A}}{1-\text{tan A}}$

$=\frac{\text{tan A}}{1-\left(\frac{1}{\text{tan A}}\right)}+\frac{\frac{1}{\text{tan A}}}{1-\text{tan A}}$

$=\frac{\text{tan A}}{\frac{\text{tan A}-1}{\text{tan A}}}+\frac{1}{\text{tan A}(1-\text{tan A})}$

$=\frac{{\text{tan}}^{2}A}{\text{tan A}-1}-\frac{1}{\text{tan A}(\text{tan A}-1)}\dots \dots (a-b=-(b-a\left)\right)$

$=\frac{{\text{tan}}^{3}A-1}{\text{tan A}(\text{tan A}-1)}$

$=\frac{\left(\text{/}\text{tan A}-1\right)({\text{tan}}^{2}A+\text{tan A}+1)}{\text{tan A}\left(\text{/}\text{tan A}-1\right)}\dots \dots [a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right]$

$=\frac{{\text{tan}}^{2}A}{\text{tan A}}+\frac{\text{tan A}}{\text{tan A}}+\frac{1}{\text{tan A}}$

$=\text{tan A + 1 + cot A}$

$=1+\frac{\text{sin A}}{cosA}+\frac{\text{cos A}}{sinA}$

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

$=1+\frac{1}{\text{sin A}\text{cos A}}$

$=1+\text{cosec A sec A}$