# Consider the following statements: 1. If y = ln (sec x + tan x), then $\rm \frac {dy}{dx} = \sec x$. 2. If y = ln (cosec x - cot x), then $\rm \fra 0 votes 13 views in Calculus closed Consider the following statements: 1. If y = ln (sec x + tan x), then \(\rm \frac {dy}{dx} = \sec x$.

2. If y = ln (cosec x - cot x), then $\rm \frac {dy}{dx} = cosec\;x$.

Which of the above is / are correct?

1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
5. None of these

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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.