Question

If A, B , C are interior angles of a triangle ABC, then find sec (B + C − A/2)

Answer 1

$\huge\orange{\underline\blue{\underline\green{\underline{ \mathcal\orange{Que}\mathcal\green{stion}}}}}❤️$

If A, B , C are interior angles of a triangle ABC, then find sec (B + C − A/2)

$\huge\orange{\underline\red{\underline\blue{\underline{ \mathcal\orange{Ans}\mathcal\green{wer}}}}}❤️$

there must be tan²(B+C)/2 instead of sec²(B+C)/2

we know that

A+B+C=180° (angle sum property)

=>B+C =180°-A

=> (B+C)/2= (180°-A)/2. [dividing both sides by 2]

we get,

=> (B+C)/2=90°-A/2

=> sec(B+C)/2=sec(90°--A/2)

=>sec(B+C)/2=cosecA/2

squaring both the sides

=> sec²(B+C)/2= cosec²A/2

=> 1+ tan²(B+C)/2= 1+cot²A/2

=> tan²(B+C)/2 = cot²A/2

LHS=RHS