Prove that \[\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}\]
Answer
363.6k+ views
Hint: Here R. H. S indicates a term which includes $\sec \theta $ so, we divide the numerator and denominator of L. H. S by $\cos \theta $.
Dividing numerator and denominator by $\cos \theta $ makes the L. H. S term easy. As $\sec \theta $ is reciprocal of $\cos \theta $ so we are dividing L .H .S by $\cos \theta $.
$ \Rightarrow $L. H. S = \[\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} \div \dfrac{{\cos \theta }}{{\cos \theta }}\]
$ \Rightarrow $L. H. S =$\dfrac{{\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}}}{{\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}}}$ ……. (1)
Now, by using trigonometric identities we will simplify the L. H. S term.
We know that
$\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $, $\dfrac{{\cos \theta }}{{\cos \theta }} = 1$, $\dfrac{1}{{\cos \theta }} = \sec \theta $
Now, putting these values in equation (1), we get
$ \Rightarrow $L. H. S = $\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}$ ……. (2)
Now, using the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$, equation (2) becomes simple. We replace the value of 1 only in the denominator to get the desired result.
Replacing value of 1 in denominator, we get
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + ({{\sec }^2}\theta - {{\tan }^2}\theta )}}$
We know that $({a^2} - {b^2}) = (a - b)(a + b)$, using this property L. H. S can be written as,
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )}}$
Taking $(\sec \theta - \tan \theta )$ common from denominator,
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{(\sec \theta - \tan \theta )( - 1 + \sec \theta + \tan \theta )}}$
Now, eliminating the same terms from the numerator and the denominator, we get
$ \Rightarrow $L. H. S =$\dfrac{1}{{\sec \theta - \tan \theta }}$= R. H. S
Hence proved.
Note: To solve such problems one should have the knowledge of trigonometric identities like ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , $se{c^2}\theta - {\tan ^2}\theta = 1$ and $\cos e{c^2}\theta - {\cot ^2}\theta = 1$ which are very useful in solving the problems. Another key point is that we have to see the R. H. S of the question to decide which identity we can use to solve the problem.
Dividing numerator and denominator by $\cos \theta $ makes the L. H. S term easy. As $\sec \theta $ is reciprocal of $\cos \theta $ so we are dividing L .H .S by $\cos \theta $.
$ \Rightarrow $L. H. S = \[\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} \div \dfrac{{\cos \theta }}{{\cos \theta }}\]
$ \Rightarrow $L. H. S =$\dfrac{{\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}}}{{\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}}}$ ……. (1)
Now, by using trigonometric identities we will simplify the L. H. S term.
We know that
$\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $, $\dfrac{{\cos \theta }}{{\cos \theta }} = 1$, $\dfrac{1}{{\cos \theta }} = \sec \theta $
Now, putting these values in equation (1), we get
$ \Rightarrow $L. H. S = $\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}$ ……. (2)
Now, using the identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$, equation (2) becomes simple. We replace the value of 1 only in the denominator to get the desired result.
Replacing value of 1 in denominator, we get
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + ({{\sec }^2}\theta - {{\tan }^2}\theta )}}$
We know that $({a^2} - {b^2}) = (a - b)(a + b)$, using this property L. H. S can be written as,
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta - \sec \theta + (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )}}$
Taking $(\sec \theta - \tan \theta )$ common from denominator,
$ \Rightarrow $L. H. S =$\dfrac{{\tan \theta - 1 + \sec \theta }}{{(\sec \theta - \tan \theta )( - 1 + \sec \theta + \tan \theta )}}$
Now, eliminating the same terms from the numerator and the denominator, we get
$ \Rightarrow $L. H. S =$\dfrac{1}{{\sec \theta - \tan \theta }}$= R. H. S
Hence proved.
Note: To solve such problems one should have the knowledge of trigonometric identities like ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , $se{c^2}\theta - {\tan ^2}\theta = 1$ and $\cos e{c^2}\theta - {\cot ^2}\theta = 1$ which are very useful in solving the problems. Another key point is that we have to see the R. H. S of the question to decide which identity we can use to solve the problem.
Last updated date: 22nd Sep 2023
•
Total views: 363.6k
•
Views today: 6.63k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

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

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE
