Question

Please solve the problem ​

Answer 1

» Question :

If $\purple{\sf{\dfrac{cot\:A}{1 + cosec\:A} - \dfrac{cot\:A}{1 - cosec\:A} = \dfrac{K}{cos\:A}}}$

then find the value of "K".

» Solution :

» To Find :

The value of "K" in the Equation.

» We Know :

• $\sf{Cot\theta = \dfrac{cos\theta}{sin\theta}}$

• $\sf{cosec\theta = \dfrac{1}{sin\theta}}$

• $\sf{(a + b)(a - b) = a^{2} - b^{2}}$

» Calculation :

$\sf{\dfrac{cot\:A}{1 + cosec\:A} - \dfrac{cot\:A}{1 - cosec\:A} = \dfrac{K}{cos\:A}}$

Using the identity,

$\sf{(a + b)(a - b) = a^{2} - b^{2}}$

We Get ,

$\sf{\Rightarrow \dfrac{cot\:A(1 - cosec\:A) - cot\:A(1 + cosec\:A}{1^{2} - cosec^{2}A}) = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \dfrac{cot\:A - cot\:Acosec\:A - cot\:A - cot\:Acosec\:A}{1 - cosec^{2}A}= \dfrac{K}{cos\:A}}$

Substituting the values of cot A and cosec A , in the Equation, we get :

$\sf{\Rightarrow \dfrac{\cancel{\dfrac{cos\:A}{sin\:A}} - \left(\dfrac{cos\:A}{sin\:A}\right)\left(\dfrac{1}{sin\:A}\right) - \cancel{\dfrac{cos\:A}{sin\:A}} - \left(\dfrac{cos\:A}{sin\:A}\right)\left(\dfrac{1}{sin\:A}\right)}{\left(1 - \dfrac{1}{sin^{2}A}\right)}= \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \dfrac{-\left(\dfrac{cos\:A}{sin\:A}\right)\left(\dfrac{1}{sin\:A}\right) - \left(\dfrac{cos\:A}{sin\:A}\right)\left(\dfrac{1}{sin\:A}\right)}{\left(1 - \dfrac{1}{sin^{2}A}\right)} = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \dfrac{-\left(\dfrac{cos\:A}{sin^{2}A}\right) - \left(\dfrac{cos\:A}{sin^{2}A}\right)}{\left(1 - \dfrac{1}{sin^{2}A}\right)} = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \dfrac{\left(\dfrac{- cos\:A - cos\:A}{sin^{2}A}\right)}{\left(\dfrac{sin^{2}A - 1}{sin^{2}A}\right)} = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \dfrac{\left(\dfrac{-2cos\:A}{sin^{2}A}\right)}{\left(\dfrac{sin^{2}A - 1}{sin^{2}A}\right)} = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \left(\dfrac{-2cos\:A}{\cancel{sin^{2}A}}\right) \times \left(\dfrac{\cancel{sin^{2}A}}{sin^{2}A - 1}\right) = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \left(\dfrac{-2cos\:A}{1}\right) \times \left(\dfrac{1}{sin^{2}A - 1}\right) = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \left(\dfrac{-2cos\:A}{sin^{2}A - 1}\right) = \dfrac{K}{cos\:A}}$

$\sf{\Rightarrow \left(\dfrac{-2cos\:A}{sin^{2}A - 1}\right) \times cos\:A = K}$

$\sf{\Rightarrow \left(\dfrac{-2cos^{2}A}{sin^{2}A - 1}\right) = K}$

$\purple{\sf{\therefore K = \left(\dfrac{-2cos^{2}A}{sin^{2}A - 1}\right)}}$

Hence ,the value of "k" is $\sf{\left(\dfrac{-2cos^{2}A}{sin^{2}A - 1}\right)}$

» Additional information :

• sin(a + b) = sinAcosB + cosAsinB

• sin(a - b) = sinAcosB - cosAsinB

• cos(a + b) = cosAcosB - sinAsinB

• cos(a - b) = cosAcosB + sinAsinB

Answer 2

GIVEN:-

$\frac{ \cot A}{1 + \cosec A} - \frac{ \cot A}{1 - \cosec A} = \frac{k}{ \cos A}$

FIND:-

k = ?

SOLUTION:-

L.H.S

$\frac{ \cot A}{1 + \cosec A} - \frac{ \cot A}{1 - \cosec A}$

now we take common:-

$\cot A (\frac{1} {1 + \cosec A} - \frac{ 1}{1 - \cosec A} )$

$\cot A (\frac{1 - \cosec A - 1 - \cosec A} { (1 + \cosec A) (1 + \cosec A)})$

$\cot A (\frac{ \cancel1 - \cosec A \cancel{ - 1} - \cosec A} { (1 + \cosec A) (1 + \cosec A)})$

we know that:-

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

so,

$\cot A (\frac{ - 2 \cosec A } { {1}^{2} - \cosec{}^{2} A) }) = \cot A (\frac{ - 2 \cosec A } { 1 - \cosec{}^{2} A })......(i)$

now, we know that

$1 + \cot {}^{2} \theta = \cosec{}^{2} \theta \\ 1 - \cosec{}^{2} \theta = - \cot {}^{2} \theta$

so,

$1 - \cosec{}^{2} A = - \cot {}^{2} A$

put this value in eq(i) we get

$\cot A (\frac{ - 2 \cosec A } { - \cot {}^{2} A}) = \cancel{ \cot A} (\frac{ \cancel - 2 \cosec A } { \cancel{ - \cot A} \cancel- \cot A})......(ii)$

now,

$\cosec A = \frac{1}{ \sin A } \\ \cot A = \frac{ \cos A }{ \sin A}$

put these values in eq(ii)

$= 2 \times \frac{ \frac{1}{\sin A} }{ \frac{\cos A }{ \sin A} }$

$= \frac{2}{ \cancel{ \cos A}} = \frac{k}{ \cancel{ \cos A} }$

$= > k = 2$

$\boxed{ so \: answer \: is \: k = 2}$