Question

tan A+secA=3find the value of sina​

Answer 1

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$If \ tan A+secA=3 \ find \ the \ value\ of\\ sinA$

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$\boxed{\red{\bold{Given:}}}$

$\purple {\implies tan A + secA =3}$

$\purple{\implies}\displaystyle \frac{sinA}{cosA} + \frac{1}{cosA} =3$

$\blue{\left(\because tanA = \displaystyle \frac{sinA}{cosA} \: and \: secA = \frac{1}{cosA} \right)}$

$\purple{\implies} \displaystyle \frac{1+sinA}{cosA} =3$

$\boxed{\red{\bold{Squaring\ both\ sides:}}}$

$\red{\implies \displaystyle \left( \frac{1+sinA}{cosA} \right)^2 = 3^2}$

$\red{\implies}\displaystyle \frac{1+sin^2 A +2 sinA}{cos^2A} = 9$

$\red{\implies}\displaystyle \frac{1+sin^2 A+ 2sinA}{1-sin^2A} = 9$

$\blue{(\because cos^2 A + sin^2A = 1)}$

$\boxed{\red{\bold{By \ cross \ multiply:}}}$

$\green{\implies 1+sin^2 A +2sinA=9(1-sin^2 A)}$

$\green{\implies}1 + sin^2 A + 2 sinA = 9 - 9 sin^2 A$

$\green{\implies}sin^2 A + 9 sin^2 A +2 sinA + 1 -9 = 0$

$\green{\implies}10 sin^2 A + 2 sinA - 8=0$

$\green{\implies}5 sin^2 A + sinA -4=0$

$\boxed{\red{\bold{Solving \ quadratic \ equation:}}}$

$\pink{\implies}5 sin^2 A + 5 sinA - 4sinA -4=0$

$\pink{\implies}5 sinA(sinA +1) -4(sinA + 1)=0$

$\pink{\implies}(sin A +1)(5 sinA -4)= 0$

$\green{\boxed{\implies{\boxed{sin A = -1 \: or \: sinA = \displaystyle \frac{4}{5}}}}}$

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