Answer
Verified
414.6k+ views
Hint: We will begin with the left hand side of the equation (1) and then we will multiply the numerator and denominator of the term in the left hand side of the expression by \[\sec A-\tan A\] and finally we will get the answer which will be equal to the right hand side of the expression. We will use the formula \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] which will help us in proving the given expression.
Complete step-by-step answer:
It is mentioned in the question that \[\dfrac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{{\tan }^{2}}A........(1)\]
Now beginning with the left hand side of the equation (1) we get,
\[\Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}........(2)\]
Now multiplying the numerator and denominator in equation (2) by \[\sec A-\tan A\] and hence we get,
\[\begin{align}
& \Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}\times \dfrac{\sec A-\tan A}{\sec A-\tan A} \\
& \Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{{{\sec }^{2}}A-{{\tan }^{2}}A}..........(3) \\
\end{align}\]
Now we know that \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]. So applying this in equation (3) we get,
\[\Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{1}..........(4)\]
Now expanding the term in equation (4) using \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] we get,
\[\Rightarrow {{\sec }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A..........(5)\]
Now again substituting \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\] in equation (5) we get,
\[\begin{align}
& \Rightarrow 1+{{\tan }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A \\
& \Rightarrow 1+2{{\tan }^{2}}A-2\sec A\tan A......(6) \\
\end{align}\]
Hence from equation (6) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.
Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (2) but here the key is to multiply the numerator and denominator of the term in the left hand side of the expression by \[\sec A-\tan A\]. Also we will use \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] a few times to get the left hand side equal to the right hand side of the equation (1).
Complete step-by-step answer:
It is mentioned in the question that \[\dfrac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{{\tan }^{2}}A........(1)\]
Now beginning with the left hand side of the equation (1) we get,
\[\Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}........(2)\]
Now multiplying the numerator and denominator in equation (2) by \[\sec A-\tan A\] and hence we get,
\[\begin{align}
& \Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}\times \dfrac{\sec A-\tan A}{\sec A-\tan A} \\
& \Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{{{\sec }^{2}}A-{{\tan }^{2}}A}..........(3) \\
\end{align}\]
Now we know that \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]. So applying this in equation (3) we get,
\[\Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{1}..........(4)\]
Now expanding the term in equation (4) using \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] we get,
\[\Rightarrow {{\sec }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A..........(5)\]
Now again substituting \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\] in equation (5) we get,
\[\begin{align}
& \Rightarrow 1+{{\tan }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A \\
& \Rightarrow 1+2{{\tan }^{2}}A-2\sec A\tan A......(6) \\
\end{align}\]
Hence from equation (6) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.
Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (2) but here the key is to multiply the numerator and denominator of the term in the left hand side of the expression by \[\sec A-\tan A\]. Also we will use \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] a few times to get the left hand side equal to the right hand side of the equation (1).
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE