If P,Q and R are interior angles of a ΔPQR , then tan(P+Q2)=

P+Q+R=180° P+Q=180°−R P+Q2=180°−R2 P+Q2=90°−R2 tan(P+Q2)=tan(90°−R2) =cot(R2)

If P,Q and R are interior angles of a ΔPQR , then tan(P+Q2)=

P+Q+R=180° P+Q=180°−R P+Q2=180°−R2 P+Q2=90°−R2 tan(P+Q2)=tan(90°−R2) =cot(R2)

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If $P, Q and R$ are interior angles of a $Δ P Q R$ , then $\tan (\frac{P + Q}{2}) =$

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\sin (\frac{R}{2})

\cos (\frac{R}{2})

\cot (\frac{R}{2})

\tan (\frac{R}{2})

answer is C.

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Detailed Solution

P + Q + R = 180 °

$P + Q = 180 ° - R$

$\frac{P + Q}{2} = \frac{180 ° - R}{2}$

$\frac{P + Q}{2} = 90 ° - \frac{R}{2}$

$\tan (\frac{P + Q}{2}) = \tan (90 ° - \frac{R}{2})$

$= \cot (\frac{R}{2})$

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If P,Q and R are interior angles of a ΔPQR , then tan(P+Q2)=

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If $P, Q and R$ are interior angles of a $Δ P Q R$ , then $\tan (\frac{P + Q}{2}) =$