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

How do you verify (sin⁡θ−1)(tan⁡θ+sec⁡θ)=−cos⁡θ(\sin \theta - 1)(\tan \theta + \sec \theta ) = - \cos \theta ?

Explanation

Solution

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⁡2x+sin⁡2x=1{\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⁡θ−1)(tan⁡θ+sec⁡θ)(\sin \theta - 1)(\tan \theta + \sec \theta )
⇒LHS=(sin⁡θ−1)(tan⁡θ+sec⁡θ)\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⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} and sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }}
So, substitute these values above.
⇒LHS=(sin⁡θ−1)(sin⁡θcos⁡θ+1cos⁡θ)\Rightarrow LHS = (\sin \theta - 1)\left( {\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)
⇒LHS=(sin⁡θ−1)(sin⁡θ+1cos⁡θ)\Rightarrow LHS = (\sin \theta - 1)\left( {\dfrac{{\sin \theta + 1}}{{\cos \theta }}} \right)
We know the algebraic identity (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}
So, using this, we can write,
⇒LHS=sin⁡2θ−12cos⁡θ\Rightarrow LHS = \dfrac{{{{\sin }^2}\theta - {1^2}}}{{\cos \theta }}
We also know a standard identity cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
And from this, we can conclude that, sin⁡2θ−1=−cos⁡2θ{\sin ^2}\theta - 1 = - {\cos ^2}\theta
Now, substitute this value in above.
⇒LHS=−cos⁡2θcos⁡θ\Rightarrow LHS = \dfrac{{ - {{\cos }^2}\theta }}{{\cos \theta }}
⇒LHS=−cos⁡θ=RHS\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 θ\theta . So, if you substitute any value of θ\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⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x)