CBSE 2026 30/8/1 · MATH · Medium · Trigonometry · SA · 3 marks
Prove that : (cscA−sinA)(secA−cosA)=1tanA+cotA(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}(cscA−sinA)(secA−cosA)=tanA+cotA1
LHS=(1sinA−sinA)(1cosA−cosA)=1−sin2AsinA⋅1−cos2AcosA=cos2A⋅sin2AsinAcosA=sinAcosA\text{LHS} = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) = \frac{1 - \sin^2 A}{\sin A} \cdot \frac{1 - \cos^2 A}{\cos A} = \frac{\cos^2 A \cdot \sin^2 A}{\sin A \cos A} = \sin A \cos ALHS=(sinA1−sinA)(cosA1−cosA)=sinA1−sin2A⋅cosA1−cos2A=sinAcosAcos2A⋅sin2A=sinAcosA. RHS=1sinAcosA+cosAsinA=1sin2A+cos2AsinAcosA=sinAcosA\text{RHS} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \sin A \cos ARHS=cosAsinA+sinAcosA1=sinAcosAsin2A+cos2A1=sinAcosA. LHS=RHS\text{LHS} = \text{RHS}LHS=RHS.
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