Question

1. Prove (1-cosA) ( 1-secA) = tanA sinA

2. Prove secA (1-sinA)(secA+tanA ) = 1

Answer 1

Answer on Question #46412 – Math – Trigonometry

Problem

1. Prove $(1 - \cos A)(1 - \sec A) = \tan A \sin A$

2. Prove $\sec A(1 - \sin A)(\sec A + \tan A) = 1$

Solution

1. $(1 - \cos A)(1 - \sec A) = \tan A \sin A$

You know that

- $\sec A = \frac{1}{\cos A}$

- $\sin^2 A + \cos^2 A = 1$

- $\tan A = \frac{\sin A}{\cos A}$

Rewrite the left-hand side

$$(1 - \cos A)(1 - \sec A) = (1 - \cos A) \left(1 - \frac{1}{\cos A}\right) = (1 - \cos A) \frac{(\cos A - 1)}{\cos A}$$

Rewrite the right-hand side

$$\tan A \sin A = \frac{\sin A}{\cos A} \sin A = \frac{\sin^2 A}{\cos A} = \frac{1 - \cos^2 A}{\cos A} = \frac{(1 - \cos A)(1 + \cos A)}{\cos A}$$

So we can see that

$$(1 - \cos A) \frac{(\cos A - 1)}{\cos A} \neq \frac{(1 - \cos A)(1 + \cos A)}{\cos A}$$

If you change the sign “-” on the “+” sign in the second bracket then you will get the correct statement $(1 - \cos A)(1 + \sec A) = \tan A \sin A$.

2. $\sec A(1 - \sin A)(\sec A + \tan A) = 1$

Rewrite the left-hand side

$$\begin{aligned} \sec A(1 - \sin A)(\sec A + \tan A) &= \frac{1}{\cos A} (1 - \sin A) \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) = \\ &= \frac{1}{\cos A} (1 - \sin A) \left(\frac{1 + \sin A}{\cos A}\right) = \frac{1}{\cos^2 A} (1 - \sin^2 A) = \frac{\cos^2 A}{\cos^2 A} = 1 \end{aligned}$$

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