Question

8. I sec A+ tanA=2/3
.then find sec A- tanA​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\large{\underline{\underline{\mathfrak{\green{\sf{Queation:-}}}}}}$.

• If sec A+ tan A=2/3,then find sec A- tanA

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$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• $\:(\sec A\:-\tan A)\:=\:1$....(1)

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$\large{\underline{\underline{\mathfrak{\sf{Find\:Here:-}}}}}$.

• $\:(\sec A\:+\tan A)\:=\:?$

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.$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

We know that ,

• $\large\boxed{\boxed{\sec^2 A - \tan^2 A \:=\:1}}$

• $\:(x^2-y^2)\:=\:(x+y)(x-y)$

So,

$\implies\:(\sec A \:+\tan A)(\sec A\:-\tan A)\:=\:1$

Keep value by (1) ,

$\implies\:(\sec A \:-\tan A)\:(\frac{2}{3})\:=\:1$

$\implies\:(\sec A \:-\tan A)\:=\frac{3}{2}$

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$\large{\underline{\underline{\mathfrak{\green{\sf{ANSWER:-}}}}}}$.

$\large\boxed{\boxed{(\sec A\:-\tan A)\:=\frac{3}{2}}}$

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