If secB +tana=P then find the value of sinB in terms of 'P'
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Answer:
SinB=(p2−1)/(p2+1)
Step-by-step explanation:
secA =p- tanA
squaring both sides: we get
sec2A=(p−tanA)2
sec2A=p2+tan2A−2ptanA
sec2A−tan2A=p2−2ptanA ,since we know the identity( 1+tan2A=secA)
hence, 1= p2−2ptanA
tanA= (p2−1)/2p
now,according to right angled triangle, tanA= perpendicular(P)/base(b)
and sinA=perpendicular(P)/hypotnuese(h)
hence we know. P= p2−1,base=2p
so, H= p2+1 (calculated by hypotnuese theorem)
now SinB= P/H= ( p2−1)/(p2+1)−−ans
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