if tan0=11/3 evaluate√sec0+cosec0/sec0-cosec0
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tan0=11/3
we know that tan0=P/B
by this we obtain that
P=11 and B=3
now we have perpendicular and base and we required hypotenuse
by applying Pythagoras we obtain H=√130
after that you put the values and solve
we know that tan0=P/B
by this we obtain that
P=11 and B=3
now we have perpendicular and base and we required hypotenuse
by applying Pythagoras we obtain H=√130
after that you put the values and solve
salmaan786:
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