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Implicit Differentiation

If tan #160...

Question

If $\tan θ = \frac{12}{13}$ , find the value of $\frac{2 \sin θ \cos θ}{\cos^{2} θ - \sin^{2} θ}$

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Solution

Given: ……(1)
To find the value of
Now, we know the following trigonometric identity
Therefore, by substituting the value of from equation (1) ,
We get,
By taking L.C.M. on the R.H.S,
We get,
Therefore
Therefore
…… (2)
Now, we know that
Therefore,
Therefore
…… (3)
Now, we know the following trigonometric identity
Therefore,
Now by substituting the value of from equation (3)
We get,
Therefore, by taking L.C.M on R.H.S
We get,
Now, by taking square root on both sides
We get,
Therefore,
…… (4)
Substituting the value of and from equation (3) and (4) respectively in the expression below
Therefore,
Therefore,

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {cos}^{2} θ & {sin}^{2} θ & 4 sin 4 θ {cos}^{2} θ & 1 + {sin}^{2} θ & 4 sin 4 θ {cos}^{2} θ & {sin}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

If the value of

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c o s^{2} θ + 2 s i n^{2} θ + 3 c o s^{2} θ + 4 s i n^{2} θ + . . . . . . (200) t e r m s = 10025

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Q. The value of

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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t a n x = \frac{2 b}{a - c} (a \neq c), y = a c o s^{2} x + 2 b s i n x c o s x + c s i n^{2} x a n d z = a s i n^{2} x - 2 b s i n x c o s x + c o s^{2} x,

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