Question

is A,B and C are the interior angles of a triangle , prove that
sec A-B+C / 2 = cosec B​

Answer 1

Here A, B and C are interior angles of a triangle.

Since the sum of interior angles of any triangle is 180°, we have,

$\longrightarrow\sf{A+B+C=180^o}$

Or,

$\longrightarrow\sf{A+C=180^o-B\quad\quad\dots(1)}$

So,

$\longrightarrow\sf{\sec\left(\dfrac{A-B+C}{2}\right)=\sec\left(\dfrac{A+C-B}{2}\right)}$

From (1),

$\longrightarrow\sf{\sec\left(\dfrac{A-B+C}{2}\right)=\sec\left(\dfrac{180^o-B-B}{2}\right)}$

$\longrightarrow\sf{\sec\left(\dfrac{A-B+C}{2}\right)=\sec\left(\dfrac{180^o-2B}{2}\right)}$

$\longrightarrow\sf{\sec\left(\dfrac{A-B+C}{2}\right)=\sec\left(90^o-B\right)}$

But,

• $\sf{\sec(90^o-x)=\csc x}$

Then,

$\longrightarrow\sf{\underline{\underline{\sec\left(\dfrac{A-B+C}{2}\right)=\csc B}}}$

Hence Proved!

Answer 2

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