Question

I am a beginner at this question . So please explain the answer clearly .

Answer 1

Proof:-

• This is the required proof.

LHS:-

$\sf 3 \sin( \frac{\pi}{6} ) \sec( \frac{\pi}{3} ) - 4 \sin( \frac{5\pi}{6} ) - \cot( \frac{\pi}{4} )$

$\sf = 3 \times \frac{1}{ \cancel{2}} \times \cancel{2} - 4 \sin(\pi - \frac{\pi}{6} ) \times 1$

$\sf = 3 - 4\sin( \frac{\pi}{6} )$

$\sf = 3 - 4 \times \frac{1}{2}$

$\sf = 3 - 2$

$\sf = 1$

RHS:-

$\sf = 1$

Hence, LHS=RHS (Hence Proved).

Answer 2

Answer:

This is the required proof.

LHS:-

\sf 3 \sin( \frac{\pi}{6} ) \sec( \frac{\pi}{3} ) - 4 \sin( \frac{5\pi}{6} ) - \cot( \frac{\pi}{4} )3sin(

6

π

)sec(

3

π

)−4sin(

6

5π

)−cot(

4

π

)

\sf = 3 \times \frac{1}{ \cancel{2}} \times \cancel{2} - 4 \sin(\pi - \frac{\pi}{6} ) \times 1=3×

2

1

×

2

−4sin(π−

6

π

)×1

\sf = 3 - 4\sin( \frac{\pi}{6} )=3−4sin(

6

π

)

\sf = 3 - 4 \times \frac{1}{2}=3−4×

2

1

\sf = 3 - 2=3−2

\sf = 1=1

RHS:-

\sf = 1=1

Hence, LHS=RHS (Hence Proved).