Answer

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Hint: Here, first of all, simplify LHS by substituting \[\cot A=\dfrac{1}{\tan A}\] everywhere. Use formulas \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] and \[\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\] to simplify LHS. Finally, substitute \[\tan A=\dfrac{\sin A}{\cos A}\] to get the answer.

Complete step-by-step answer:

Here, we have to show that \[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A\]

Let us consider the left-hand side (LHS) of the given expression as

\[E=\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}\]

First of all, we take the LCM of denominators, we get,

\[E=\dfrac{\tan A\left( 1-\tan A \right)+\cot A\left( 1-\cot A \right)}{\left( 1-\cot A \right)\left( 1-\tan A \right)}\]

Now, by simplifying the above expression, we get

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{{{1}^{2}}-\tan A-\cot A+\cot A.\tan A}\]

We know that, \[\tan A=\dfrac{1}{\cot A}\]

By multiplying cot A on both sides, we get,

\[\cot A.\tan A=\dfrac{\cot A}{\cot A}\]

Hence, \[\cot A.\tan A=1\]

By applying this in the above expression, we get,

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{1-\tan A-\cot A+1}\]

By further simplifying the above expression, we get,

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{2-\tan A-\cot A}\]

Now, by substituting \[\cot A=\dfrac{1}{\tan A}\] in the above expression, we get,

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\dfrac{1}{\tan A}-\dfrac{1}{{{\tan }^{2}}A}}{2-\tan A-\dfrac{1}{\tan A}}\]

By simplifying the above expression, we get,

\[E=\dfrac{\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{{{\tan }^{2}}A}}{\dfrac{2\tan A-{{\tan }^{2}}A-1}{\tan A}}\]

By cancelling the like terms, we get,

\[E=\dfrac{\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{\tan A}}{\dfrac{2\tan A-{{\tan }^{2}}A-1}{1}}\]

We can also write the above expression as,

\[E=\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{\left( \tan A \right)\left( 2\tan A-{{\tan }^{2}}A-1 \right)}\]

By multiplying –1 on both numerator and denominator of the above expression, we get,

\[E=\dfrac{{{\tan }^{4}}A-{{\tan }^{3}}A-\tan A+1}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}\]

We can also write the above expression as,

\[E=\dfrac{{{\tan }^{3}}A\left( \tan A-1 \right)-1\left( \tan A-1 \right)}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}\]

By taking (tan A – 1) common in the numerator, we get,

\[E=\dfrac{\left( {{\tan }^{3}}A-1 \right)\left( \tan A-1 \right)}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}\]

We know that \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]. By applying this in the denominator, we get,

\[E=\dfrac{\left( {{\tan }^{3}}A-1 \right)\left( \tan A-1 \right)}{\left( \tan A \right){{\left( \tan A-1 \right)}^{2}}}\]

By cancelling the like terms, we get

\[E=\dfrac{\left( {{\tan }^{3}}A-{{1}^{3}} \right)}{\left( \tan A \right)\left( \tan A-1 \right)}\]

We know that \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]. By applying this in the numerator, we get,

\[E=\dfrac{\left( \tan A-1 \right)\left( {{\tan }^{2}}A+\tan A+1 \right)}{\left( \tan A \right)\left( \tan A-1 \right)}\]

By cancelling the like terms, we get

\[E=\dfrac{\left( {{\tan }^{2}}A \right)+\left( \tan A \right)+1}{\left( \tan A \right)}\]

We can also write the above expression as,

\[E=\dfrac{{{\tan }^{2}}A}{\tan A}+\dfrac{\tan A}{\tan A}+\dfrac{1}{\tan A}\]

By cancelling the like terms, we get,

\[E=\tan A+1+\dfrac{1}{\tan A}\]

By substituting\[\tan A=\dfrac{\sin A}{\cos A}\], we get

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

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

By simplifying the above expression, we get,

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

We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. By applying this in the above expression, we get,

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

Since we know that \[\sin A=\dfrac{1}{\operatorname{cosec}A}\] and \[\cos A=\dfrac{1}{\sec A}\]. We get,

\[E=\dfrac{1}{\dfrac{1}{\operatorname{cosec}A}.\dfrac{1}{\sec A}}+1\]

Therefore, we get,

\[E=\operatorname{cosec}A.\sec A+1=LHS\]

Therefore, E = LHS = RHS

Hence, we have shown that \[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A\]

Note: Students can also solve this question by substituting \[\tan A=\dfrac{\sin A}{\cos A}\] and \[\cot A=\dfrac{\cos A}{\sin A}\] in the first step only. Students must take special care while taking \[{{\tan }^{2}}A\] as LCM in the numerator as they often make mistakes while writing the powers of tan A after taking the LCM. Also, students should always try to resolve the expression into factors by using identities like \[{{\left( a\pm b \right)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab\] \[\left( {{a}^{3}}\pm {{b}^{3}} \right)=\left( a\pm b \right)\left( {{a}^{2}}\mp ab+{{b}^{2}} \right)\] etc.

Complete step-by-step answer:

Here, we have to show that \[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A\]

Let us consider the left-hand side (LHS) of the given expression as

\[E=\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}\]

First of all, we take the LCM of denominators, we get,

\[E=\dfrac{\tan A\left( 1-\tan A \right)+\cot A\left( 1-\cot A \right)}{\left( 1-\cot A \right)\left( 1-\tan A \right)}\]

Now, by simplifying the above expression, we get

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{{{1}^{2}}-\tan A-\cot A+\cot A.\tan A}\]

We know that, \[\tan A=\dfrac{1}{\cot A}\]

By multiplying cot A on both sides, we get,

\[\cot A.\tan A=\dfrac{\cot A}{\cot A}\]

Hence, \[\cot A.\tan A=1\]

By applying this in the above expression, we get,

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{1-\tan A-\cot A+1}\]

By further simplifying the above expression, we get,

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\cot A-{{\cot }^{2}}A}{2-\tan A-\cot A}\]

Now, by substituting \[\cot A=\dfrac{1}{\tan A}\] in the above expression, we get,

\[E=\dfrac{\tan A-{{\tan }^{2}}A+\dfrac{1}{\tan A}-\dfrac{1}{{{\tan }^{2}}A}}{2-\tan A-\dfrac{1}{\tan A}}\]

By simplifying the above expression, we get,

\[E=\dfrac{\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{{{\tan }^{2}}A}}{\dfrac{2\tan A-{{\tan }^{2}}A-1}{\tan A}}\]

By cancelling the like terms, we get,

\[E=\dfrac{\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{\tan A}}{\dfrac{2\tan A-{{\tan }^{2}}A-1}{1}}\]

We can also write the above expression as,

\[E=\dfrac{{{\tan }^{3}}A-{{\tan }^{4}}A+\tan A-1}{\left( \tan A \right)\left( 2\tan A-{{\tan }^{2}}A-1 \right)}\]

By multiplying –1 on both numerator and denominator of the above expression, we get,

\[E=\dfrac{{{\tan }^{4}}A-{{\tan }^{3}}A-\tan A+1}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}\]

We can also write the above expression as,

\[E=\dfrac{{{\tan }^{3}}A\left( \tan A-1 \right)-1\left( \tan A-1 \right)}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}\]

By taking (tan A – 1) common in the numerator, we get,

\[E=\dfrac{\left( {{\tan }^{3}}A-1 \right)\left( \tan A-1 \right)}{\left( \tan A \right)\left( {{\tan }^{2}}A-2\tan A+1 \right)}\]

We know that \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]. By applying this in the denominator, we get,

\[E=\dfrac{\left( {{\tan }^{3}}A-1 \right)\left( \tan A-1 \right)}{\left( \tan A \right){{\left( \tan A-1 \right)}^{2}}}\]

By cancelling the like terms, we get

\[E=\dfrac{\left( {{\tan }^{3}}A-{{1}^{3}} \right)}{\left( \tan A \right)\left( \tan A-1 \right)}\]

We know that \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]. By applying this in the numerator, we get,

\[E=\dfrac{\left( \tan A-1 \right)\left( {{\tan }^{2}}A+\tan A+1 \right)}{\left( \tan A \right)\left( \tan A-1 \right)}\]

By cancelling the like terms, we get

\[E=\dfrac{\left( {{\tan }^{2}}A \right)+\left( \tan A \right)+1}{\left( \tan A \right)}\]

We can also write the above expression as,

\[E=\dfrac{{{\tan }^{2}}A}{\tan A}+\dfrac{\tan A}{\tan A}+\dfrac{1}{\tan A}\]

By cancelling the like terms, we get,

\[E=\tan A+1+\dfrac{1}{\tan A}\]

By substituting\[\tan A=\dfrac{\sin A}{\cos A}\], we get

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

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

By simplifying the above expression, we get,

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

We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. By applying this in the above expression, we get,

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

Since we know that \[\sin A=\dfrac{1}{\operatorname{cosec}A}\] and \[\cos A=\dfrac{1}{\sec A}\]. We get,

\[E=\dfrac{1}{\dfrac{1}{\operatorname{cosec}A}.\dfrac{1}{\sec A}}+1\]

Therefore, we get,

\[E=\operatorname{cosec}A.\sec A+1=LHS\]

Therefore, E = LHS = RHS

Hence, we have shown that \[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A\]

Note: Students can also solve this question by substituting \[\tan A=\dfrac{\sin A}{\cos A}\] and \[\cot A=\dfrac{\cos A}{\sin A}\] in the first step only. Students must take special care while taking \[{{\tan }^{2}}A\] as LCM in the numerator as they often make mistakes while writing the powers of tan A after taking the LCM. Also, students should always try to resolve the expression into factors by using identities like \[{{\left( a\pm b \right)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab\] \[\left( {{a}^{3}}\pm {{b}^{3}} \right)=\left( a\pm b \right)\left( {{a}^{2}}\mp ab+{{b}^{2}} \right)\] etc.

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