Question
Prove that: 3 sin $\frac { \pi } { 6 }$ sec $\frac { \pi } { 3 }$ - 4 sin $\frac { 5 \pi } { 6 }$ cot $\frac { \pi } { 4 }$ = 1.
✓
Answer
LHS $= 3 sin $ $\frac { \pi } { 6 }$ $\times$$ sec$ $\frac { \pi } { 3 }$ $- 4 sin$ $\frac { 5 \pi } { 6 }$ $\times$ $cot$ $\frac { \pi } { 4 }$
= 3 $\times$ $\frac { 1 } { 2 }$ $\times$ 2 - 4 sin $\left( \pi - \frac { \pi } { 6 } \right)$ $\times$ cot $\frac { \pi } { 4 }$
= 3 - 4 sin $\frac { \pi } { 6 }$ $\times$ 1 [$\because$ sin $( \pi - \theta )$ = sin $\theta$]
$= 3 - 4$ $\times$ $\frac { 1 } { 2 }$ $= 3 - 2 = 1 =$ RHS
$\therefore$ LHS = RHS
Hence proved.
= 3 $\times$ $\frac { 1 } { 2 }$ $\times$ 2 - 4 sin $\left( \pi - \frac { \pi } { 6 } \right)$ $\times$ cot $\frac { \pi } { 4 }$
= 3 - 4 sin $\frac { \pi } { 6 }$ $\times$ 1 [$\because$ sin $( \pi - \theta )$ = sin $\theta$]
$= 3 - 4$ $\times$ $\frac { 1 } { 2 }$ $= 3 - 2 = 1 =$ RHS
$\therefore$ LHS = RHS
Hence proved.
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