If A,B,C are interior angles of triangle ABC .show that sec square (B+C/2)=cot square (A/2)
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Answered by
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hello there !!!
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
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
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My tuition teacher has given
Previous years question paper
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No
It is right
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I am hundreds percent sure
Okay
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Answered by
42
Answer:I have done it in the simplest way known by me... It's actually been taught by my father ....
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