Question

How do you verify \[(\sin \theta - 1)(\tan \theta + \sec \theta ) = - \cos \theta \]?

Answer 1

Hint: We use trigonometric identities and ratios to solve this problem. We use some methods of simplifying algebraic expressions and verify the equation by equating the left-hand side and right-hand side of the given equation.
The formulas that we use in this problem are \[{\cos ^2}x + {\sin ^2}x = 1\] which is a standard trigonometric identity.

Complete step by step solution:
To solve this problem, consider the left-hand side of the equation and the right-hand side of the equation. And then we simplify the left-hand side of the equation and check whether we are getting the same expression present on the right-hand side as our result.
Now, on the left hand side of the equation, it is given as \[(\sin \theta - 1)(\tan \theta + \sec \theta )\]
\[ \Rightarrow LHS = (\sin \theta - 1)(\tan \theta + \sec \theta )\]
We know that, tangent of an angle is the ratio of sine of that angle to cosine of that angle i.e., \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\sec \theta = \dfrac{1}{{\cos \theta }}\]
So, substitute these values above.
\[ \Rightarrow LHS = (\sin \theta - 1)\left( {\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)\]
\[ \Rightarrow LHS = (\sin \theta - 1)\left( {\dfrac{{\sin \theta + 1}}{{\cos \theta }}} \right)\]
We know the algebraic identity \[(a - b)(a + b) = {a^2} - {b^2}\]
So, using this, we can write,
\[ \Rightarrow LHS = \dfrac{{{{\sin }^2}\theta - {1^2}}}{{\cos \theta }}\]
We also know a standard identity \[{\cos ^2}\theta + {\sin ^2}\theta = 1\]
And from this, we can conclude that, \[{\sin ^2}\theta - 1 = - {\cos ^2}\theta \]
Now, substitute this value in above.
\[ \Rightarrow LHS = \dfrac{{ - {{\cos }^2}\theta }}{{\cos \theta }}\]
\[ \Rightarrow LHS = - \cos \theta = RHS\]
Hence, we proved that the Left-hand side is equal to the Right-hand side.

Note:
As we have verified this equation, this equation is now true for any value of \[\theta \]. So, if you substitute any value of \[\theta \] in Left-hand side and Right-hand side, you will get equal values.
While simplifying expressions, try to change them in such a way that you get any known identities, or you get terms that are already present in the expression, so that you get rid of those complicated terms. Here, we converted some terms and we got a known expression. (\[{\cos ^2}x = 1 - {\sin ^2}x\])