Prove the following trigonometric equation:
\[\dfrac{\tan A}{1-\cot A}+\dfrac{\cot A}{1-\tan A}=1+\sec A\operatorname{cosec}A\]
Last updated date: 21st Mar 2023
•
Total views: 305.4k
•
Views today: 3.84k
Answer
305.4k+ views
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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
