Question

#QualityQuestion
@Trigonometry

$\sf{If~ tan~ (cot~ x) ~= ~cot~ (tan ~x), ~then~ sin ~2x ~is?}$​

Answer 1

~Solution :-

• Here, we can,

$\sf{tan (cotx) = cot (tanx) }$

$\to \sf{ tan (cotx) = tan ( \frac{π }{2 } - tanx)}$

$\to \sf{cotx = nπ + \frac{π}{ 2} - tanx}$

$\to \sf{⇒ cotx + tanx = nπ + \frac{π}{ 2}}$

$\to \sf{2 sin2x = nπ + \frac{ π }{ 2}}$

$\to \sf{ sin2x = [ \frac{2}{ nπ }+ { \frac{π }{2}}]}$

$</p><p> \implies \bf \red{= \frac{ 4 }{ {(2n + 1) π}}}$

• Hence, 2x is valued to $\sf { \frac{ 4 }{ {(2n + 1) π}}}$.

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Answer 2

Answer:

~Solution :-

Here, we can,

\sf{tan (cotx) = cot (tanx) }tan(cotx)=cot(tanx)

\begin{gathered} \to \sf{ tan (cotx) = tan ( \frac{π }{2 } - tanx)} \\ \end{gathered}

→tan(cotx)=tan(

2

π

−tanx)

\begin{gathered} \to \sf{cotx = nπ + \frac{π}{ 2} - tanx} \\ \end{gathered}

→cotx=nπ+

2

π

−tanx

\begin{gathered} \to \sf{⇒ cotx + tanx = nπ + \frac{π}{ 2}} \\ \end{gathered}

→⇒cotx+tanx=nπ+

2

π

\begin{gathered} \to \sf{2 sin2x = nπ + \frac{ π }{ 2}} \\ \end{gathered}

→2sin2x=nπ+

2

π

\begin{gathered} \to \sf{ sin2x = [ \frac{2}{ nπ }+ { \frac{π }{2}}]} \\ \end{gathered}

→sin2x=[

nπ

2

+

2

π

]

\begin{gathered} < /p > < p > \implies \bf \red{= \frac{ 4 }{ {(2n + 1) π}}} \\ \end{gathered}