if tan A =6 find the value of sec A
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tanA = p/b=6/1
Let p=6x, b=1x
Then by Pythagoras theorem
H=^\/p2+b2=^\/36x2+1x2=^\/37 x
Now secA=h/b=^\/37 x/1x=^\/37
Let p=6x, b=1x
Then by Pythagoras theorem
H=^\/p2+b2=^\/36x2+1x2=^\/37 x
Now secA=h/b=^\/37 x/1x=^\/37
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tanA=6=6/1=p/b
(p×p)+(b×b)= (h×h)
(6×6)+(1×1)= (h×h)
36+1=(h×h)
√37= h
secA= h/b =√37/1= √37
(p×p)+(b×b)= (h×h)
(6×6)+(1×1)= (h×h)
36+1=(h×h)
√37= h
secA= h/b =√37/1= √37
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