If 3tan A = 4 then prove that :(i) √secA-cosecA/√secA+cosecA = 1/√3 (ii) 1−sinA/ 1+cos A = 1 /2√ 2
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3tanA=4
or, tanA=4/3=p/b
Using Pythagorus theorem, h²=p²+b² we get,
h²=4²+3²
or, h²=16+9
or, h²=25
or, h=5 (neglecting the negative sign)
∴, secA=h/b=5/3 and cosecA=h/p=5/4
i) √secA-cosecA/√secA+cosecA
=√(5/3-5/4)/(5/3+5/4)
=√(5/12)/√(35/12)
=√(5/12×12/35)
=√1/7
ii) 1-sinA/1+cosA
=(1-4/5)/(1+3/5)
={(5-4)/5}/{(5+3)/5}
=(1/5)/(8/5)
=1/5×5/8
=1/8
or, tanA=4/3=p/b
Using Pythagorus theorem, h²=p²+b² we get,
h²=4²+3²
or, h²=16+9
or, h²=25
or, h=5 (neglecting the negative sign)
∴, secA=h/b=5/3 and cosecA=h/p=5/4
i) √secA-cosecA/√secA+cosecA
=√(5/3-5/4)/(5/3+5/4)
=√(5/12)/√(35/12)
=√(5/12×12/35)
=√1/7
ii) 1-sinA/1+cosA
=(1-4/5)/(1+3/5)
={(5-4)/5}/{(5+3)/5}
=(1/5)/(8/5)
=1/5×5/8
=1/8
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