Cos A-1-sin A-1-sin A-cos A-2 A

The correct option is A True cos A/(1+sin A)+ (1+sin A)/cos A = 2 sec A L.H.S. = cos A/(1+sin A)+ (1+sin A)/cos A = [cos^2A+ (1+sin A)^2]/(1+sin A)cos A = ( ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios

cos A/1+sin A...

Question

$\frac{c o s A}{(1 + s i n A)} + \frac{(1 + s i n A)}{c o s A} = 2 s e c A$

True

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False

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Solution

The correct option is A
True

$\frac{c o s A}{(1 + s i n A)} + \frac{(1 + s i n A)}{c o s A} = 2 s e c A$
$L . H . S . = \frac{c o s A}{(1 + s i n A)} + \frac{(1 + s i n A)}{c o s A} = \frac{[c o s^{2} A + (1 + s i n A)^{2}]}{(1 + s i n A) c o s A} = \frac{(c o s^{2} A + s i n^{2} A + 1 + 2 s i n A)}{(1 + s i n A) c o s A} = \frac{(1 + 1 + 2 s i n A)}{(1 + s i n A) c o s A} = \frac{(2 + 2 s i n A)}{(1 + s i n A) c o s A} = \frac{2 (1 + s i n A)}{(1 + s i n A) c o s A} = \frac{2}{c o s A} = 2 s e c A = R . H . S .$

Suggest Corrections

cosA(1+sinA)+(1+sinA)cosA=2secA

The correct option is A True cosA(1+sinA)+(1+sinA)cosA=2secA L.H.S.=cosA(1+sinA)+(1+sinA)cosA=[cos2A+(1+sinA)2](1+sinA)cosA=(cos2A+sin2A+1+2sinA)(1+sinA)cosA=(1+1+2sinA)(1+sinA)cosA=(2+2sinA)(1+sinA)cosA=2(1+sinA)(1+sinA)cosA=2cosA=2secA=R.H.S.

$\frac{c o s A}{(1 + s i n A)} + \frac{(1 + s i n A)}{c o s A} = 2 s e c A$