Question
Without using trigonometric table, prove that
$\cos ^2 26^{\circ}+\cos 64^{\circ} \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}}=2$
$\cos ^2 26^{\circ}+\cos 64^{\circ} \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}}=2$
✓
Answer
$\text { LHS }=\cos ^2 26^{\circ}+\cos 64^{\circ} \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot 54^{\circ}}$
$=\cos ^2 26^{\circ}+\cos \left(90^{\circ}-26^{\circ}\right) \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-54^{\circ}\right)}$
$=\cos ^2 26^{\circ}+\sin 26^{\circ} \cdot \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\tan 36^{\circ}}$
$=\cos ^2 26^{\circ}+\sin ^2 26+1 \ldots .\left(\cos ^2 \theta+\sin ^2 \theta=1\right)$
$=1+1=2$
$=\text { RHS }$
Hence proved.
$=\cos ^2 26^{\circ}+\cos \left(90^{\circ}-26^{\circ}\right) \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-54^{\circ}\right)}$
$=\cos ^2 26^{\circ}+\sin 26^{\circ} \cdot \sin 26^{\circ}+\frac{\tan 36^{\circ}}{\tan 36^{\circ}}$
$=\cos ^2 26^{\circ}+\sin ^2 26+1 \ldots .\left(\cos ^2 \theta+\sin ^2 \theta=1\right)$
$=1+1=2$
$=\text { RHS }$
Hence proved.
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