Question

$\huge\sf{Question}$
Prove (sec A + tan A)(1-sin A) =cos A​

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Answer 1

Step-by-step explanation:

(a + b)(a – b)= a2 – b2,

So, (1 + sin A)(1 – sin A) = 1 – sin2A.

Hence, the value of (sec A + tan A) (1 – sin A) is equal to cos A.

Answer 2

Given :-

• (sec A + tan A)(1-sin A) =cos A

Solution:-

Taking LHS -

$\begin{gathered}\\\quad\longrightarrow\quad \sf (sec \: A + tan \: A)(1-sin \: A) \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf (\frac{1}{cos\:A} + \frac{sin\:A}{cos \:A})(1-sin \:A) \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf (\frac{1+sin\:A}{cos\:A} )(1-sin \:A) \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf \frac{1^2-sin^2\:A}{cos\:A} \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf \frac{1-sin^2\:A}{cos\:A} \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf \frac{1-sin^2\:A}{cos\:A} \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf \frac{cos^2\:A}{cos\:A} \\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad \sf cos\:A\\\end{gathered}$

=RHS

Hence Proved.