QQuestion
Prove that:
\frac{(\cot \mathrm{A}+\tan \mathrm{A}-1)(\sin \mathrm{A}+\cos \mathrm{A})}{\sin^3 \mathrm{A}+\cos^3 \mathrm{A}}=\sec \mathrm{A}\times\cosec \mathrm{A}
ICSEClass 10MathematicsTrigonometrical Identities04 Mark2025 Hard
✓Answer
Starting with the left-hand side:
\frac{(\cot \mathrm{A}+\tan \mathrm{A}-1)(\sin \mathrm{A}+\cos \mathrm{A})}{\sin^3 \mathrm{A}+\cos^3 \mathrm{A}} =\frac{\left(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}+\frac{\sin \mathrm{A}}{\cos \mathrm{A}}-1\right)(\sin \mathrm{A}+\cos \mathrm{A})}{(\sin \mathrm{A}+\cos \mathrm{A})(\sin^2 \mathrm{A}-\sin \mathrm{A}\cos \mathrm{A}+\cos^2 \mathrm{A})}Using sin² A + cos² A = 1:
=\frac{\left(\frac{\cos^2 \mathrm{A}+\sin^2 \mathrm{A}}{\sin \mathrm{A}\cos \mathrm{A}}-1\right)(\sin \mathrm{A}+\cos \mathrm{A})}{(\sin \mathrm{A}+\cos \mathrm{A})(1-\sin \mathrm{A}\cos \mathrm{A})} =\frac{\left(\frac{1-\sin \mathrm{A}\cos \mathrm{A}}{\sin \mathrm{A}\cos \mathrm{A}}\right)(\sin \mathrm{A}+\cos \mathrm{A})}{(\sin \mathrm{A}+\cos \mathrm{A})(1-\sin \mathrm{A}\cos \mathrm{A})}Cancelling common factors:
=\frac{1}{\sin \mathrm{A}\cos \mathrm{A}} =\frac{1}{\cos \mathrm{A}}\times\frac{1}{\sin \mathrm{A}}= sec A × cosec A
Therefore, LHS = RHS. Hence proved.
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