Prove the given trigonometric expression: \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\tan A+\sec A\]
Answer
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Hint: Simplify the left hand side of a given trigonometric expression by multiplying and dividing it by \[\tan A+\sec A-1\]. Rearrange the terms and simplify the equation using the identity \[{{\tan }^{2}}A+1={{\sec }^{2}}A\] to prove the given expression.
Complete step-by-step answer:
We have to prove the trigonometric expression \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\tan A+\sec A\]. We will simplify the left hand side of the given equation to equate it to the right hand side.
So, we will multiply and divide \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}\] by \[\tan A-\sec A-1\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}\times \dfrac{\tan A-\sec A-1}{\tan A-\sec A-1}\].
We know the identity \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\].
Substituting \[a=\tan A-\sec A,b=1\] in the above expression, we have \[\left( \tan A-\sec A+1 \right)\left( \tan A-\sec A-1 \right)={{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}\].
Substituting \[a=\tan A-1,b=\sec A\] in the above expression, we have \[\left( \tan A+\sec A-1 \right)\left( \tan A-\sec A-1 \right)={{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}\times \dfrac{\tan A+\sec A-1}{\tan A+\sec A-1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}\].
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{{{\tan }^{2}}A+1-2\tan A-{{\sec }^{2}}A}{{{\tan }^{2}}A+{{\sec }^{2}}A-2\tan A\sec A-1}\].
We know that \[{{\tan }^{2}}A+1={{\sec }^{2}}A\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{{{\sec }^{2}}A-2\tan A-{{\sec }^{2}}A}{{{\tan }^{2}}A+1+{{\tan }^{2}}A-2\tan A\sec A-1}\].
Further simplifying the above expression, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{-2\tan A}{2{{\tan }^{2}}A-2\tan A\sec A}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{\tan A}{\tan A\sec A-{{\tan }^{2}}A}\].
Factorizing the denominator of the above expression, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{\tan A}{\tan A\left( \sec A-\tan A \right)}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{1}{\sec A-\tan A}\].
Multiplying and dividing the above expression by \[\tan A+\sec A\], we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1}{\sec A-\tan A}=\dfrac{1}{\sec A-\tan A}\times \dfrac{\tan A+\sec A}{\tan A+\sec A}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1}{\sec A-\tan A}=\dfrac{\tan A+\sec A}{{{\sec }^{2}}A-{{\tan }^{2}}A}\] using the identity \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\].
We know that \[{{\tan }^{2}}A+1={{\sec }^{2}}A\]. Thus, we have \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\].
So, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1}{\sec A-\tan A}=\dfrac{\tan A+\sec A}{{{\sec }^{2}}A-{{\tan }^{2}}A}=\dfrac{\tan A+\sec A}{1}=\tan A+\sec A\].
Hence, we have proved that \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\tan A+\sec A\].
Note: Usual mistakes while solving trigonometry are-
(i) Error in applying identities.
(ii) Not able to comprehend how to take the next step.
(iii) Starting the simplification of both sides simultaneously. (When solving problems you can either take the LHS or the RHS. Observe and analyze which way it is easy to arrive at the other side Use techniques such as substitution, factorisation, rationalisation, create common denominators, accordingly to simplify and attain the other side.)
Complete step-by-step answer:
We have to prove the trigonometric expression \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\tan A+\sec A\]. We will simplify the left hand side of the given equation to equate it to the right hand side.
So, we will multiply and divide \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}\] by \[\tan A-\sec A-1\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}\times \dfrac{\tan A-\sec A-1}{\tan A-\sec A-1}\].
We know the identity \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\].
Substituting \[a=\tan A-\sec A,b=1\] in the above expression, we have \[\left( \tan A-\sec A+1 \right)\left( \tan A-\sec A-1 \right)={{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}\].
Substituting \[a=\tan A-1,b=\sec A\] in the above expression, we have \[\left( \tan A+\sec A-1 \right)\left( \tan A-\sec A-1 \right)={{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}\times \dfrac{\tan A+\sec A-1}{\tan A+\sec A-1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}\].
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{{{\tan }^{2}}A+1-2\tan A-{{\sec }^{2}}A}{{{\tan }^{2}}A+{{\sec }^{2}}A-2\tan A\sec A-1}\].
We know that \[{{\tan }^{2}}A+1={{\sec }^{2}}A\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{{{\sec }^{2}}A-2\tan A-{{\sec }^{2}}A}{{{\tan }^{2}}A+1+{{\tan }^{2}}A-2\tan A\sec A-1}\].
Further simplifying the above expression, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{-2\tan A}{2{{\tan }^{2}}A-2\tan A\sec A}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{\tan A}{\tan A\sec A-{{\tan }^{2}}A}\].
Factorizing the denominator of the above expression, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{\tan A}{\tan A\left( \sec A-\tan A \right)}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{{{\left( \tan A-1 \right)}^{2}}-{{\left( \sec A \right)}^{2}}}{{{\left( \tan A-\sec A \right)}^{2}}-{{\left( 1 \right)}^{2}}}=\dfrac{1}{\sec A-\tan A}\].
Multiplying and dividing the above expression by \[\tan A+\sec A\], we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1}{\sec A-\tan A}=\dfrac{1}{\sec A-\tan A}\times \dfrac{\tan A+\sec A}{\tan A+\sec A}\].
Thus, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1}{\sec A-\tan A}=\dfrac{\tan A+\sec A}{{{\sec }^{2}}A-{{\tan }^{2}}A}\] using the identity \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\].
We know that \[{{\tan }^{2}}A+1={{\sec }^{2}}A\]. Thus, we have \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\].
So, we have \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\dfrac{1}{\sec A-\tan A}=\dfrac{\tan A+\sec A}{{{\sec }^{2}}A-{{\tan }^{2}}A}=\dfrac{\tan A+\sec A}{1}=\tan A+\sec A\].
Hence, we have proved that \[\dfrac{\tan A+\sec A-1}{\tan A-\sec A+1}=\tan A+\sec A\].
Note: Usual mistakes while solving trigonometry are-
(i) Error in applying identities.
(ii) Not able to comprehend how to take the next step.
(iii) Starting the simplification of both sides simultaneously. (When solving problems you can either take the LHS or the RHS. Observe and analyze which way it is easy to arrive at the other side Use techniques such as substitution, factorisation, rationalisation, create common denominators, accordingly to simplify and attain the other side.)
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