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# Using the trigonometric identity ${\sec ^2}\theta = 1 + {\tan ^2}\theta$, Prove that $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}$.

Last updated date: 20th Jun 2024
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Hint: In this question, we have to equate the given particular of the left-hand side with the right-hand side.
We need to first put the trigonometric formulas to bring it in a shorter form so that we can get the solution, then taking common from numerator we will get a shorter form than putting the algebraic and trigonometric formulas we can prove L.H.S equals to R.H.S.

Formula used:
${\sec ^2}\theta = 1 + {\tan ^2}\theta$.
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$

Complete step by step answer:
We need to prove that, $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}$.
We will use the formula, ${\sec ^2}\theta = 1 + {\tan ^2}\theta$.
Now,
L.H.S.= $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}}$
Dividing numerator and denominator by $\cos \theta$, we get,
$\Rightarrow \dfrac{{\dfrac{{\sin \theta - \cos \theta + 1}}{{\cos \theta }}}}{{\dfrac{{\sin \theta + \cos \theta - 1}}{{\cos \theta }}}}$
Cancelling the similar terms in numerator and denominator we get,
$\Rightarrow \dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }} - 1 + \dfrac{1}{{\cos \theta }}}}{{\dfrac{{\sin \theta }}{{\cos \theta }} + 1 - \dfrac{1}{{\cos \theta }}}}$
We know that, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }},\sec \theta = \dfrac{1}{{\cos \theta }}$ , using these we get,
$\Rightarrow \dfrac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}$
Multiply both numerator and denominator by $\left( {\tan \theta - \sec \theta } \right)$
$\Rightarrow \dfrac{{\left( {\tan \theta + \sec \theta - 1} \right)\left( {\tan \theta - \sec \theta } \right)}}{{\left( {\tan \theta - \sec \theta + 1} \right)\left( {\tan \theta - \sec \theta } \right)}}$
Solving we get,
$\Rightarrow \dfrac{{\left( {\tan \theta + \sec \theta } \right)\left( {\tan \theta - \sec \theta } \right) - 1\left( {\tan \theta - \sec \theta } \right)}}{{\left( {\tan \theta - \sec \theta + 1} \right)\left( {\tan \theta - \sec \theta } \right)}}$
Using the algebraic formula, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$\Rightarrow \dfrac{{\left( {{{\tan }^2}\theta - {{\sec }^2}\theta } \right) - \tan \theta + \sec \theta }}{{\left( {\tan \theta - \sec \theta + 1} \right)\left( {\tan \theta - \sec \theta } \right)}}$
Using the trigonometric formula, ${\sec ^2}\theta = 1 + {\tan ^2}\theta$
$\Rightarrow \dfrac{{ - 1 - \tan \theta + \sec \theta }}{{\left( {\tan \theta - \sec \theta + 1} \right)\left( {\tan \theta - \sec \theta } \right)}}$
Taking $- 1$ common from numerator, we get,
$\Rightarrow \dfrac{{ - \left( {1 + \tan \theta - \sec \theta } \right)}}{{\left( {\tan \theta - \sec \theta + 1} \right)\left( {\tan \theta - \sec \theta } \right)}}$
Cancelling the same term from numerator and denominator, we get,
$\Rightarrow \dfrac{{ - 1}}{{\tan \theta - \sec \theta }}$
Simplifying we get,
$\Rightarrow \dfrac{1}{{\sec \theta - \tan \theta }}$
=R.H.S.
$\therefore$ L.H.S. = R.H.S.

Hence, $\dfrac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \dfrac{1}{{\sec \theta - \tan \theta }}$. Hence proved.

Note:
We have to focus on the calculations. Here we use so many values and trigonometry relations in this problem so we have to mind that we concentrate on doing the calculations. Sin Cos formulas are based on sides of the right-angled triangle. Sin and Cos are basic trigonometric functions along with tan functions in trigonometry. Sine of angle is equal to the ratio of opposite side and hypotenuse whereas cosine of an angle is equal to ratio of adjacent side and hypotenuse.
$\sin \theta = \dfrac{{Opposite \, side}}{{Hypotenuse}} \\ \cos \theta = \dfrac{{Adjacent\, side}}{{Hypotenuse}} \\$

In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. The most widely used trigonometric functions are the sine, the cosine, and the tangent.