Question

1+tan²A upon 1+secA =sec A
prove
pls solve step by step
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Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

How does one solve 1+ (tan^2A) \ (1+secA) = secA?

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

To prove 1+tan²A upon 1+secA =sec A

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

You want us to solve this: This means you want us to find the value{s) of A that satisfy this equation.

1+tan2(A)1+sec(A)=sec(A)

Multiplying both sides of the equation by [1 + sec(A)]

1+sec(A)+tan2(A)=sec(A)+sec2(A)

Now we’ll use one of the basic trigonometric identities,

sec2(A)=tan2(A)+1→tan2(A)=sec2(A)−1

So,

1+sec(A)+sec2(A)−1=sec(A)+sec2(A)

sec(A)+sec2(A)=sec(A)+sec2(A)

So, what values of A satisfy this equation?

Answer: Every number.

You worded the question incorrectly; you should have asked us to prove that the equation was true.

1+tan2(A)1+sec(A)

=1+sec(A)+tan2(A)1+sec(A)

=1+sec(A)+sec2(A)−11+sec(A)

=sec(A)[1+sec(A)]1+sec(A)

=sec(A)

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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