Question

If sinQ=3/5FIND THE VALUE OF Sec²Q+tan²Q

Answer 1

Step-by-step explanation:

by pythagerous theorm its sides are (3,4,5,)

secQ=5/4,tan=3/4

sec^2Q+tan^2Q

(5/4)^2+(3/4)^2

25/16+9/16

34/16

Answer 2

Question :

$\sf{If\:sinQ=\frac{3}{5},find\: the\: value\:of\:sec^2Q+tan^2Q.}$

Solution :

$\sf{sinQ=\frac{3}{5}}$

We know,

${\boxed{\bold{sinA=\frac{Height}{Hypotenuse}}}}$

Here,

• Height = 3
• Hypotenuse = 5

✪ Now find the base by using ‘ Pythagoras Theorem '✪

$\sf{Base^2+Height^2=Hypotenuse^2}$

$\implies\sf{Base^2+3^2=5^2}$

$\implies\sf{Base^2+9=25}$

$\implies\sf{Base^2=25-9}$

$\implies\sf{Base=\sqrt{16}}$

$\implies\sf{Base=4}$

• Base = 4

We know,

★${\boxed{\bold{secA=\frac{Hypotenuse}{Base}}}}$

★${\boxed{\bold{tanA=\frac{Height}{Base}}}}$

• $\sf{secQ=\frac{5}{4}}$

• $\sf{tanQ=\frac{3}{4}}$

✪ Now find the value of sec²Q + tan²Q ✪

$\sf{sec^2Q+tan^2Q}$

$\sf{Put\:secQ=\frac{5}{4}\:\: and\:\:tanQ=\frac{3}{4}}$

$\sf{=(\frac{5}{4})^2+(\frac{3}{4})^2}$

$\sf{=(\frac{25}{16})+(\frac{9}{16})}$

$\sf{=\frac{25+9}{16}}$

$\sf{=\frac{34}{16}}$

$\sf{=\frac{17}{8}}$

Therefore, the value of sec²Q + tan²Q is 17/8.

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Some formulas related to trigonometry :-

• sin²A + cos²A = 1

• 1+tan²A = sec²A

• 1+cot²A = cosec²A

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