Question

Tan theta/1-cot theta +cot theta/1-tan theta= 1+sec theta cosec theta

Answer 1

To prove

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$\frac{ \tan(x) }{1 - \cot(x) } + \frac{ \cot(x) }{1 - \tan(x) } = 1 + \sec(x) . \csc(x)$

Solution

Let theta = x

✪ While solving this type of questions we take on one side and prove it equivalent to the other given side , so applying this method firstly we take LHS and prove it equivalent to RHS.

LHS

$\frac{ \tan(x) }{1 - \cot(x ) } + \frac{ \cot(x) }{1 - \tan(x) }$

As we know that, we can write

tan x = sinx/cos x

Cot x = cos x / sin x

So replacing tan and cot from above mentioned values:

$\frac{ \frac{ \sin(x) }{ \cos(x) } }{1 - \frac{ \cos(x) }{ \sin(x) } } + \frac{ \frac{ \cos(x) }{ \sin(x) } }{1 - \frac{ \sin(x) }{ \cos(x) } }$

Now we will perform LCM of the digits on denominator place of both fractions.

$\frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{ \sin(x) - \cos(x) ) }{ \sin(x) } } + \frac{ \frac{ \cos(x) }{ \sin(x) } }{ \frac{ \cos(x) - \sin(x) ) }{ \cos(x) } }$

$\frac{ { \sin(x) }^{2} }{ \cos(x)( \sin(x ) - \cos(x) ) } + \frac{ \cos( {x}^{2} ) }{ \sin(x) ( \cos(x) - \sin(x) }$

Now we will take ( - ) as common from our 2nd fraction then the positive values will change into negative values, applying this we get :

$\frac{ { \sin(x) }^{2} }{ \cos(x)( \sin(x) - \cos(x) )} - \frac{ { \cos(x) }^{2} }{ \sin(x)( \sin(x) - \cos(x) )}$

(as we took ( - ) as common)

Now again we will take the LCM of these two fractions, then we will get :

$\frac{ { \sin(x) }^{3} - { \cos(x) }^{3} }{ \sin(x) . \cos(x) ( \sin(x) - \cos(x) ) }$

We will apply here the algebraic identity (a³-b³ = (a-b)(a²+b² +ab)

$\frac{( \sin(x) - \cos(x) )( { \sin(x) }^{2} + { \cos(x) }^{2} + \sin(x). \cos(x) }{ \sin(x). \cos(x) ( \sin(x) - \cos(x) }$

Here sin X - cos x is present in both numerator and denominator so it will get cancelled.

and as we know that

sin²x + cos²x = 1

So we get :

$\frac{1 + \sin(x). \cos(x) }{ \sin(x). \cos(x) }$

$\frac{1}{ \sin(x) . \cos(x) } + \frac{ \sin(x). \cos(x) }{ \sin(x). \cos(x) }$

As in 2nd fraction numerator and denominator are same so they will also cancelled.

$\frac{1}{ \sin(x) } \times \frac{1}{ \cos(x) } + 1$

As we know that

1/sin X = cosec X

1/cos x = sec x

So replacing the values with sec and cosec we get

$\csc(x) . \sec(x) + 1$

or

LHS =1 + sec x . cosec X= RHS

Hence proved

Answer 2

GIVEN:-

$\sf \frac{ \tan \theta}{1 - cot\theta } + \frac{ \cot \theta}{ 1 - \tan \theta} = 1 + \sec \theta \cosec \theta$

TO FIND:-

$\tt lhs = rhs$

SOLUTION:-

$\tt lhs \ratio -$

$\sf \frac{ \tan \theta}{1 - cot\theta } + \frac{ \cot \theta}{ 1 - \tan \theta}$

$\tt = \frac{ \frac{ \sin \theta }{ \cos\theta } }{1 - \frac{\cos\theta}{\sin \theta} } + \frac{\frac{\cos\theta}{\sin \theta} }{1 - \frac{ \sin \theta }{ \cos\theta }}$

$\tt = \frac{ \frac{ \sin \theta }{ \cos\theta } }{ \frac{ \sin \theta- \cos\theta}{ \sin \theta} } + \frac{\frac{\cos\theta}{\sin \theta} }{ \frac{ \cos\theta - \sin \theta }{ \cos\theta }}$

$\tt = \frac{ \sin \theta }{ \cos \theta } \times \frac{ \sin \theta }{(\sin \theta - \cos \theta ) } + \frac{\cos \theta }{\sin \theta} \times \frac{ \cos \theta }{( \cos \theta - \sin \theta ) }$

$\tt = \frac{ { \sin}^{2}\theta}{ \cos \theta( \sin \theta - \cos \theta) } - \frac{ { \cos }^{2} \theta }{ \sin \theta( \sin \theta - \cos \theta ) }$

$\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{ { \sin}^{2} \theta }{ \cos \theta } - \frac{ { \cos }^{2} \theta}{ \sin \theta } ]$

$\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{ { \sin}^{3 } \theta - { \cos }^{3} \theta }{ \sin \theta \cos \theta } ]$

$\tt = \frac{1}{( \sin \theta - \cos \theta) } [ \frac{( \sin \theta - \cos \theta)( { \sin}^{2} \theta + { \cos }^{2} \theta + \sin \theta \cos \theta )}{ \sin \theta \cos \theta } ]$

$\tt = \frac{(1 + \sin \theta \cos \theta)}{( \sin \theta \cos \theta)} = \frac{1}{ \sin \theta \cos \theta } + \frac{( \sin \theta \cos \theta) }{( \sin \theta \cos \theta) }$

$= \sec \theta \cosec \theta + 1$

$= rhs$