Question

sin x / 1+ cos x -1+ cos x / sinx = 4 find x​

Answer 1

GIVEN :–

$\\ \implies\bf \dfrac{ \sin(x) }{1 + \cos(x) } - \dfrac{1 + \cos(x) }{ \sin(x) } = 4$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \implies\bf \dfrac{ \sin(x) }{1 + \cos(x) } - \dfrac{1 + \cos(x) }{ \sin(x) } = 4$

$\\ \implies\bf \dfrac{ \sin(x) . \sin(x) - \{ 1 + \cos(x)\}^{2} }{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ \sin^{2}(x)- \{ 1 + \cos(x)\}^{2} }{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ \sin^{2}(x)- \{ 1 + \cos^{2} (x) + 2 \cos(x) \}}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ \sin^{2}(x)- 1 - \cos^{2} (x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{1 - \cos^{2}(x)- 1 - \cos^{2} (x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ - \cos^{2}(x) - \cos^{2} (x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ - 2\cos^{2}(x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ - 2 \cos(x) \{ 1 + \cos(x) \}}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4$

$\\ \implies\bf \dfrac{ - 2 \cos(x)}{\sin(x)}= 4$

$\\ \implies\bf \dfrac{ \cos(x)}{\sin(x)}= - 2$

$\\ \implies\bf \dfrac{ \sin(x)}{\cos(x)}= \dfrac{1}{ - 2}$

$\\ \implies\bf \tan(x) = - \dfrac{1}{ 2}$

$\\ \implies\bf x = \tan^{ - 1} \bigg(- \dfrac{1}{ 2} \bigg)$

$\\ \large \implies{\boxed{\bf x = - \tan^{ - 1} \bigg( \dfrac{1}{ 2} \bigg)}}$

Answer 2

$\huge \mathfrak{ \underline{ \orange{ \: given}}}$

$\begin{gathered}\\ \bf \dfrac{ \sin(x) }{1 + \cos(x) } - \dfrac{1 + \cos(x) }{ \sin(x) } = 4 \\\end{gathered}$

$\huge \mathfrak{ \underline{ \pink{ \: \: Find}}} :$

- • Value of 'x' = ?

$\huge \mathfrak{ \underline{ \purple{ \: solution}}}$

$\begin{gathered}\\ \\ \dfrac{ \sin(x) }{1 + \cos(x) } - \dfrac{1 + \cos(x) }{ \sin(x) } = 4 \\\end{gathered}$

$\begin{gathered}\\ \ \dfrac{ \sin(x) . \sin(x) - \{ 1 + \cos(x)\}^{2} }{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p></p><p>\begin{gathered}\\ \ \dfrac{ \sin^{2}(x)- \{ 1 + \cos(x)\}^{2} }{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p></p><p>\begin{gathered}\\ \ \dfrac{ \sin^{2}(x)- \{ 1 + \cos^{2} (x) + 2 \cos(x) \}}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \ \dfrac{ \sin^{2}(x)- 1 - \cos^{2} (x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \\dfrac{1 - \cos^{2}(x)- 1 - \cos^{2} (x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p> </p><p></p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \\dfrac{ - \cos^{2}(x) - \cos^{2} (x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \ \dfrac{ - 2\cos^{2}(x) - 2 \cos(x)}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \ \dfrac{ - 2 \cos(x) \{ 1 + \cos(x) \}}{ \{1 + \cos(x) \} \{ \sin(x) \}}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p></p><p>\begin{gathered}\\ \\dfrac{ - 2 \cos(x)}{\sin(x)}= 4 \\\end{gathered} \\ </p><p></p><p> </p><p></p><p>\begin{gathered}\\ \i \dfrac{ \cos(x)}{\sin(x)}= - 2 \\\end{gathered} \\ </p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \ \dfrac{ \sin(x)}{\cos(x)}= \dfrac{1}{ - 2} \\\end{gathered} \\ </p><p></p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \\tan(x) = - \dfrac{1}{ 2} \\\end{gathered} \\ </p><p></p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \x = \tan^{ - 1} \bigg(- \dfrac{1}{ 2} \bigg) \\\end{gathered} </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\\ \large \i{\boxed{\bf x = - \tan^{ - 1} \bigg( \dfrac{1}{ 2} \bigg)}} \\\end{gathered} \\ </p><p></p><p> </p><p>$