Cosec A-sin A A-cos A-1-tan A- A

The correct option is A True (cosec A sin A)(sec A cos A) = 1/(tan A + cot A) L.H.S = (cosec A sin A)(sec A cos A) = (1/sin A sin A)(1/cos A cos A ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

cosec A-sin A...

Question

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

True

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False

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Solution

The correct option is A
True

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

L.H.S = R.H.S

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Similar questions

Prove: $(cosec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}$

Prove (cosecA - sinA) (sec A - cos A) = $\frac{1}{t a n A + c o t A}$

(c o s e c A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

[Hint: Simplify LHS and RHS separately]

(csc A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

(c o sec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

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(cosecA−sinA)(secA−cosA)=1(tanA+cotA)

The correct option is A True (cosecA−sinA)(secA−cosA)=1(tanA+cotA) L.H.S=(cosecA−sinA)(secA−cosA)=(1sinA−sinA)(1cosA−cosA)=[(1−sin2A)sinA][(1−cos2A)cosA]=(cos2AsinA)×(sin2AcosA)=cosAsinAR.H.S=1(tanA+cotA)=1(sinAcosA+cosAsinA)=1[(sin2A+cos2A)sinAcosA]=cosAsinA\ L.H.S = R.H.S

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