Question

if \frac{\sin x}{\cos x} \times \frac{\sec x}{\operatorname{cosec} x} \times \frac{\tan x}{\cot x}
cosx
sinx
​
×
cosecx
secx
​
×
cotx
tanx
​
= 9 where x \inx∈ (0, \frac{\pi}{2}0,
2
π
​
)then the value of x is

\frac{\pi}{4}
4
π
​

\frac{\pi}{3}
3
π
​

\frac{\pi}{2}
2
π
​

\frac{\pi}{6}
6
π
​

Answer 1

Step-by-step explanation:

sorry frnd..i can't understand your question....

Answer 2

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{Appropriate‌}}}}}} \\ \end{gathered}$   $\begin{gathered}\Large{\red{\underline{\textsf{\textbf{Question}}}}} \\ \end{gathered}$

Find definite integral of :

$\begin{gathered} \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x) \\ \end{gathered}$

ln(sinx+cosx)

​${\underline{\pink{\boxed{\bf{\gray{Answer}}}}}}}}\:\green\bigstar \\ \end{gathered}$

$\\ \\ \\ \green{\tt : \implies I = - \frac{\pi}{4} (ln \: 2)}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\begin{gathered}\green{\underline{\bold{Given :}}} \\ \tt: \implies \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx =? \end{gathered}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx = I -----(1)$

$\\ \\ \green{\to} \tt \: sin \: x + cos \: x = \sqrt{2} \bigg( \frac{1}{ \sqrt{2} } sin \: x + \frac{1}{ \sqrt{2} } cos \: x) \\ \\ \green{\to} \tt \: sin \: x + cos \: x = \sqrt{2} \bigg(cos \: \frac{\pi}{4} \times sin \: x + sin \: \frac{\pi}{4} \times cos \: x\bigg)\\\\ \green{\star}\tt\:sin(A+B)=(sin\:A\:cos\:A+cos\:A\:sin\:B )\\ \\ \green{\to} \tt \: sin \: x +$

$\tt : \implies I= \int \limits_{0}^{ \frac{\pi}{2} } ln (\sqrt{2} sin \: \theta)d \theta\\ \\ \tt : \implies I= \int \limits_{0}^{ \frac{\pi}{2} }$

$( ln \sqrt{2} )d \theta + \int \limits_{0}^{ \frac{\pi}{2} } \:ln( sin \: \theta) \: d \theta \\ \\ \green{ \circ } \tt \: \int \limits_{0}^{ \frac{\pi}{2} } ln(sin \: \theta)d \theta = - \frac{\pi}{2} (ln \: 2)\\ \\ \tt : \implies I= ln( \sqrt{2}) \int \limits_{0}^{ \frac{\pi}{2} }d \theta - \frac{\pi}{2} (ln \: 2) \\ \\ \tt : \implies I= ln( \sqrt{2}) \bigg[\the$

Hence the Answer is $\\ \\ \\ \green{\tt : \implies I = - \frac{\pi}{4} (ln \: 2)}$