Correct Answer - Option 3 : Both 1 and 2
Concept:
y = ln x then \(\rm \frac{dy}{dx} = \frac{1}{x}\)
\(\rm y = tan\;x \; then\; \frac{dy}{dx} = sec^2\;x\)
\(\rm y = sec\;x \; then\; \frac{dy}{dx} = sec\;x.tan\;x\)
\(\rm y = cosec\;x \; then\; \frac{dy}{dx} = -cosec\;x.cot\;x\)
\(\rm y = cot\;x \; then\; \frac{dy}{dx} = cosec^2\;x\)
Calculations:
Let, y = ln (sec x + tan x)
⇒\(\rm\frac{dy}{dx} = \frac{1}{sec\;x+tan\;x}\times (sec\;x.tan\;x + sec^2\;x)\)
⇒\(\rm\frac{dy}{dx} = \frac{sec\;x}{sec\;x+tan\;x}\times (tan\;x + sec\;x)\)
⇒\(\rm\frac{dy}{dx} = sec\;x\)
Statement (1) is true.
Now,
y = ln (cosec x - cot x),
⇒\(\rm\frac{dy}{dx} = \frac{1}{cosec\;x-cot\;x}\times (-cosec\;x.cot\;x +co sec^2\;x)\)
⇒\(\rm\frac{dy}{dx} = \frac{cosec\;x}{cosec\;x-cot\;x}\times (cosec\;x. - cot\;x )\)
⇒\(\rm\frac{dy}{dx} = {cosec\;x}\)
Statement (2) is true.