A,B,C are interior angles of △ABC. Prove that csc(2A+B)=sec2C.

A,B,C are interior angles of △ABC⇒∠A+∠B+∠C=180∘⇒∠A+∠B=180∘−∠C⇒2∠A+∠B=2180∘−∠C⇒2∠A+∠B=90∘−2∠C⇒csc(2∠A+∠B)=csc(90∘−2∠C)∴csc(2∠A+∠B)=sec2∠CHence proved.

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A, B, C are interior angles of ABC. Prove that ({A+B}{2})={C}

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$A, B, C$ are interior angles of $△ A BC$ . Prove that $csc (\frac{A + B}{2}) = sec \frac{C}{2}$ .

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Chloe Discussed

A, B, C

are interior angles of

△ A BC

. Prove that

csc (\frac{A + B}{2}) = sec \frac{C}{2}

10 mins ago

Discuss this question LIVE

10 mins ago

A, B, C

are interior angles of

△ A BC

\Rightarrow ∠ A + ∠ B + ∠ C = 180^{\circ}

\Rightarrow ∠ A + ∠ B = 180^{\circ} - ∠ C

\Rightarrow \frac{∠ A + ∠ B}{2} = \frac{180 ^{\circ} - ∠ C}{2}

\Rightarrow \frac{∠ A + ∠ B}{2} = 90^{\circ} - \frac{∠ C}{2}

\Rightarrow csc (\frac{∠ A + ∠ B}{2}) = csc (90^{\circ} - \frac{∠ C}{2})

∴ csc (\frac{∠ A + ∠ B}{2}) = sec \frac{∠ C}{2}

Hence proved.

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Question Text	$A, B, C$ are interior angles of $△ A BC$ . Prove that $csc (\frac{A + B}{2}) = sec \frac{C}{2}$ .
Answer Type	Text solution:1
Upvotes	150