If a, b and c are interior angles of δ abc, show that sec² (b+c / 2) -1 = cot²a/2 trigonometry class 10 maths. pls me asap.

First, it would be nice if we could put all of our angle measures in terms of a single variable. Since we're dealing with the interior angles of a triangle, and since the interior angles of a triangle add to 180°, we can say that

A + B + C = 180, or equivalently,

B + C = 180 - A

Knowing this, we can replace the B + C on the left side of the equation with 180 - A so that our equation reads:

Doing a little tidying, we can rewrite the term as

By definition, , so I'd like to rewrite the left side of the equation with these last two things in mind:

, so we can rewrite the left side again as

Next, we note that - by definition, . This allows us the rewrite the right side of the equation as

And our full equation as

Next, we want to take advantage of the most essential trig identity, , specifically the form we get by subtracting from either side: . We can transform the left side of this problem's equation into the same form by multiplying either side by . Here's what we get out of that:

The identity tells us this statement must be true, so we're done at this point.

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

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