Answer

Verified

445.5k+ views

**Hint:**For proving \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A\], we divide the LHS part of the equation by $\sin A$ and convert this into $\dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}$ and now we substitute identity in place of 1 i.e. $\cos e{c^2}A - {\cot ^2}A = 1$ in numerator after this we apply an identity in the numerator i.e. ${a^2} - {b^2} = (a + b)(a - b)$ after applying this identity we get terms as $\dfrac{{(\cos ecA + \cot A)(1 - \cos ecA + \cot A)}}{{\cot A + 1 - \cos ecA}}$ and we cancel out the like terms and we got our answer which is equal to the RHS.

**Complete step-by-step answer:**

By taking LHS \[\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}}\] and we divide each term of numerator and denominator by $\sin A$.

We get,

\[\dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}\]

After solving we get the equation as,

$\dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}$

Now, in numerator in place of 1 we insert an identity which is,

$\cos e{c^2}A - {\cot ^2}A = 1$

We get the result as,

$\dfrac{{\cot A - (\cos e{c^2}A - {{\cot }^2}A) + \cos ecA}}{{\cot A + 1 - \cos ecA}}$

Now, we put an algebraic identity which is

${a^2} - {b^2} = (a + b)(a - b)$ where, $a = \cos ecA,b = \cot A$

We get,

\[\dfrac{{\cos ecA + \cot A - [(\cos ecA + \cot A)(\cos ecA - \cot A)]}}{{\cot A + 1 - \cos ecA}}\]

Now by taking $(\cos ecA + \cot A)$ we get the equation as,

\[\dfrac{{(\cos ecA + \cot A)[1 - (\cos ecA - \cot A)]}}{{\cot A + 1 - \cos ecA}}\]

By solving bracket we get

\[\]\[\dfrac{{(\cos ecA + \cot A)[\cot A + 1 - \cos ecA]}}{{\cot A + 1 - \cos ecA}}\]

By eliminating the like terms i.e. \[\cot A + 1 - \cos ecA\]

We get,

\[\cos ecA + \cot A\]

=RHS

Hence Proved

**Note:**Alternate Method to solve the above question

By taking LHS

\[\dfrac{{\cos A - (\sin A - 1)}}{{\cos A + (\sin A - 1)}}\]

And by dividing both numerator and denominator by \[\cos A + \sin A - 1\]

We get,

=\[\dfrac{{\cos A - (\sin A - 1)}}{{\cos A + (\sin A - 1)}} \times \dfrac{{\cos A + (\sin A - 1)}}{{\cos A + (\sin A - 1)}}\]

By applying identities (i) ${a^2} - {b^2} = (a + b)(a - b)$ in numerator, (ii) ${(a + b)^2} = {a^2} + {b^2} + 2ab$ in denominator

We get,

=\[\dfrac{{{{\cos }^2}A - {{(\sin A - 1)}^2}}}{{{{[\cos A + (\sin A - 1)]}^2}}}\]

=$\dfrac{{{{\cos }^2}A - {{(\sin A - 1)}^2}}}{{{{\cos }^2}A + {{(\sin A - 1)}^2} + 2\cos A \times (\sin A - 1)}}$

Now we apply identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$ on $\sin A - 1$ in both numerator and denominator

We get,

=$\dfrac{{{{\cos }^2}A - ({{\sin }^2}A + 1 - 2\sin A)}}{{{{\cos }^2}A + {{\sin }^2}A + 1 - 2\sin A + 2\cos A\sin A - 2\cos A}}$

As, \[{\cos ^2}A + {\sin ^2}A = 1\]

=\[\dfrac{{{{\cos }^2}A - {{\sin }^2}A - 1 + 2\sin A}}{{1 + 1 - 2\sin A + 2\cos A\sin A - 2\cos A}}\]

=\[\dfrac{{{{\cos }^2}A - (1 - {{\cos }^2}A) - 1 + 2\sin A}}{{2 - 2\sin A + 2\cos A\sin A - 2\cos A}}\]

=\[\dfrac{{{{\cos }^2}A - 1 + {{\cos }^2}A - 1 + 2\sin A}}{{2(1 - \sin A) + 2\cos A(\sin A - 1)}}\]

=\[\dfrac{{2{{\cos }^2}A - 2 + 2\sin A}}{{2(1 - \sin A) - 2\cos A(1 - \sin A)}}\]

Taking (-2) common from numerator and \[2(1 - \sin A)\] common from denominator

We get

=\[\dfrac{{ - 2(1 - {{\cos }^2}A - \sin A)}}{{2(1 - \sin A)(1 - \cos A)}}\]

=\[\dfrac{{ - ({{\sin }^2}A - \sin A)}}{{(1 - \sin A)(1 - \cos A)}}\]

Taking $\sin A$ common in numerator we get,

=\[\dfrac{{ - \sin A(\sin A - 1)}}{{(1 - \sin A)(1 - \cos A)}}\]

=\[\dfrac{{\sin A(1 - \sin A)}}{{(1 - \sin A)(1 - \cos A)}}\]

=\[\dfrac{{\sin A}}{{(1 - \cos A)}}\]

Now multiply and divide numerator and denominator by \[(1 + \cos A)\]

=\[\dfrac{{\sin A}}{{(1 - \cos A)}} \times \dfrac{{(1 + \cos A)}}{{(1 + \cos A)}}\]

We get

=\[\dfrac{{\sin A \times (1 + \cos A)}}{{1 - {{\cos }^2}A}}\]

=\[\dfrac{{\sin A \times (1 + \cos A)}}{{{{\sin }^2}A}}\]

We get

=\[\dfrac{{1 + \cos A}}{{\sin A}}\]

i.e. \[\dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}}\]

=$\cos ecA + \cot A$

=RHS

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

At which age domestication of animals started A Neolithic class 11 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

One cusec is equal to how many liters class 8 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE