Question

This is challenge to bring user try to solve it

If tan θ = 12/13 then,

Find

2sin θcos θ ÷ 4sin^2 θ - sin^2θ

Chapter - Trigonometric ratio. Please solve it asap.

Try to give full content.

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

${\boxed{tan\theta = \dfrac{12}{13}}}$

As we know that

${\boxed{tan\theta = \dfrac{Perpendicular}{Base}}}$

Perpendicular = 12

Base = 13

 ${\boxed{Using\; Pythagoras\; Theorem}}$

$\tt{\rightarrow H =\sqrt{(12)^2+(13)^2}}$

$\tt{\rightarrow H =\sqrt{144+169}}$

 $\tt{\rightarrow H =\sqrt{313}}$

Now

 ${\boxed{sin\theta = \dfrac{12}{\sqrt{313}}}}$

 ${\boxed{cos\theta = \dfrac{13}{\sqrt{313}}}}$

★Now substitute the values

 $\tt{\rightarrow\dfrac{2sin\theta cos\theta}{cos^2\theta - sin^2\theta}}}$

$\tt{\rightarrow\dfrac{2\times 12/(\sqrt{313}\times 13/(\sqrt{313})}{13^2/313-12^2/313}}$

$\tt{\rightarrow\dfrac{312}{169-144}}$

 ${\boxed{\dfrac{312}{25}}}$

$\boxed{\begin{minipage}{7 cm} Fundamental Trignometric Indentities \\ \\ \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$

Answer 2

Answer:

Here I have solved your answer:

Answer:

Below

Explanation:

θ

can be in the first quadrant

0≤θ≤90 or the fourth quadrant 270≤θ≤360

If θ is in the first quadrant,

then

sinθ

=5/13

cosθ

=12/13

tanθ=5/12

Therefore,

sin2θ=2sinθcosθ=2×5/13×12/13=120/169

cos2θ=cos2θ−sin2θ=(12/13)whole square−(5/13)whole square

=144/169−25)169

=119/169

If θ is in the fourth quadrant,

then

sinθ=−5/13

cosθ=12/13

tanθ=−5/12

Therefore,

sin2θ=2sinθcosθ=2×−5\13×12\13

=−120\169

cos2θ=(cos)whole squareθ−(sin)whole squareθ=(12/13)whole square−(−5/13) whole square

=144/169−25/169

=119/169

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