Question

if sec theta + tan theta minus 2, then prove that cosec theta + cot theta is equal to 3?​

Answer 1

Answer:

Let Theta = A

Sec A + Tan A= -2

As, Sec²A - Tan²A=1

$\frac{secA+tanA}{sec^2A-tan^2A}=-2\\\\ \frac{secA+tanA}{(secA-tanA)(secA+tanA)}=-2\\\\ secA-tanA=\frac{-1}{2}$

SecA +tan A+secA-tanA=-2 $-\frac{1}{2}$

secA= $-\frac{5}{4}$

SecA +tan A-secA+tanA=-2 $+\frac{1}{2}$

tanA= $-\frac{3}{4}$

$\frac{secA}{tanA}=\frac{\frac{-5}{4}}{\frac{-3}{4}}\\\\ cosecA=\frac{5}{3}$

cotA= $-\frac{4}{3}, as cotA=\frac{1}{tanA}$

But cosec A is positive in first and second quadrant , and value of cot A is negative, it means

cosecA +cotA

       $=\frac{-5}{3}+\frac{-4}{3} {\text{or}}=\frac{5}{3}+\frac{4}{3}\\\\=\frac{-9}{3} {\text{or}}=\frac{9}{3}\\\\ -3{\text{or}}3$

Hence , cosecA +cotA=3