Question

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Answer 1

||✪✪ QUESTION ✪✪||

if 4*Tan@ = 3 , find the value of [ (4sin@ - cos@ + 1) / (4sin@ + cos@ - 1) ]

|| ★★ FORMULA USED ★★ ||

• Tan@ = Perpendicular / Base
• sin@ = Perpendicular / Hypotenuse
• cos@ = Base / Hypotenuse

|| ✰✰ ANSWER ✰✰ ||

Given that,

→ 4Tan@ = 3

→ Tan@ = (3/4) = Perpendicular / Base .

So, we can say that :-

→ Perpendicular = 3

→ Base = 4

Now, By Pythagoras Theoram , we can say that :-

=> (Hypotenuse)² = (Perpendicular)² + (Base)²

So ,

→ H² = 4² + 3²

→ H² = 16 + 9

→ H² = 25

→ H = 5 .

_____________________

So, we can say that now :-

→ sin@ = Perpendicular / Hypotenuse = 3/5

→ cos@ = Base / Hypotenuse = 4/5 .

____________________

Putting These value in Question Now, we get,

→ [ (4sin@ - cos@ + 1) / (4sin@ + cos@ - 1) ]

→ [ (4 * 3/5) - (4/5) + 1 ] / [ (4 * 3/5) + (4/5) - 1 ]

→ [ 12/5 - 4/5 + 1 ] / [ 12/5 + 4/5 - 1 ]

Taking LCM ,

→ [ (12-4+5)/5 ] / [ (12+4-5)/5 ]

→ [ (13/5) / (11/5) ]

→ (13/5) * (5/11)

→ (13/11) (Ans).

So, our Answer is (13/11).

Answer 2

$\huge\star\frak{\underline{\underline{\:AnSwER}}}$

$\large{\underline{\:Given}}$

$\pink{\sf{\mapsto{\:4\tan \theta\:=\:3}}}$

$\large{\underline{\:Evaluate}}$

$\orange{\sf{\mapsto{\left(\dfrac{4\sin \theta - \cos \theta + 1}{4\sin \theta + \cos \theta - 1}\right)}}}$

we Have

$\green{\mapsto{\sf{\:\tan \theta\:=\frac{Perpendicular}{Base}}}}$

$\green{\mapsto{\sf{\:\sin \theta\:=\frac{Perpendicular}{Hypotenuse}}}}$

$\green{\mapsto{\sf{\:\cos \theta\:=\frac{Base}{Hypotenuse}}}}$

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A/C. TO PICTURE

$\:\:\:\:\green{\sf{\:\:(by\:Pythagoras\:theorem)}}$

$\bold{\pink{\boxed{\boxed{\orange{\:(Hypotenuse)^2\:=\:(Perpendicular)^2+(Base)^2}}}}}$

$\mapsto\sf{\:(AC)^2\:=\:(AB)^2+(BC)^2}$

$\mapsto\sf{\:(AC)^2\:=\:(3)^2+(4)^2}$

$\mapsto\sf{\:(AC)^2\:=\:9+16}$

$\mapsto\sf{\:(AC)\:=\:\sqrt{25}}$

$\mapsto\boxed{\boxed{\red{\sf{\:(AC)\:=\:5}}}}$

___________________________

And,

$\mapsto{\sf{\:\sin \theta \:=\frac{AB}{AC}}}$

$\red{\mapsto{\sf{\:\sin \theta \:=\frac{3}{5}}}}$ ....(1)

And,

$\mapsto{\sf{\:\cos \theta \:=\frac{BC}{AC}}}$

$\red{\mapsto{\sf{\:\cos \theta \:=\frac{4}{5}}}}$ .....(2)

_________________________

Now,

$\sf{\left(\dfrac{4\sin \theta - \cos \theta + 1}{4\sin \theta + \cos \theta - 1}\right)}$

$\:\:\:\:\red{\sf{\:(Keep\:Values\:by\:(1)\:and\:(1) )}}$

$\mapsto\sf{\left(\dfrac{4\times\dfrac{3}{5}- \dfrac{4}{5} + 1}{4\times\dfrac{3}{5} + \dfrac{4}{5}- 1}\right)}$

$\mapsto\sf{\left(\dfrac{\dfrac{12-4+5}{5}}{\dfrac{12+4-5}{5}}\right)}$

$\mapsto\sf{\left(\dfrac{\dfrac{13}{\cancel{5}}}{\dfrac{11}{\cancel{5}}}\right)}$

$\orange{\mapsto\sf{\dfrac{13}{11} AnS.}}$