plz solve this question....if u send me the pic of solution...then it will be very helpful for me...
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we know angleA + angleB + angleC = 180 degree
A+B+C=180
B+C=180-A
DIVIDE BOTH BY 2
B+C÷2=180÷2-A÷2
B+C/2=90-A/2
now in question 1: cos^2A/2 +cos B+C/2
BUT VALUE OF B+C/2 IN QUESTION 1 , WE GET
COS^2 A/2+ COS^2 (90-A/2)
= COS^2 A/2 + SIN^2 A/2
( WE NOW COS 90-THETA IS SIN THETA)
so the ans wer will be 1 beacause cos^2 theta + sin^2 theta = 1
hence prove
we know angleA + angleB + angleC = 180 degree
A+B+C=180
B+C=180-A
DIVIDE BOTH BY 2
B+C÷2=180÷2-A÷2
B+C/2=90-A/2
now in question 1: cos^2A/2 +cos B+C/2
BUT VALUE OF B+C/2 IN QUESTION 1 , WE GET
COS^2 A/2+ COS^2 (90-A/2)
= COS^2 A/2 + SIN^2 A/2
( WE NOW COS 90-THETA IS SIN THETA)
so the ans wer will be 1 beacause cos^2 theta + sin^2 theta = 1
hence prove
nitin58:
hai the quetion after interior angel not seeing
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