Question

Prove that sec 50° sin 40° + cos 40° cosec 50° = 2.​

Answer 1

Answer:

2

Step-by-step explanation:

sec 50° sin 40° + cos 40° cosec 50°

=> sec(90° - 40°) * sin 40° + cos (90° - 50°) * cosec 50°

=> cosec 40° * sin 40° + sin 50° * cosec 50°

=> cosec 40° * (1/cosec 40°) + (1/cosec 50°) * cosec 50°

=> 1 + 1

=> 2

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Answer 2

To Prove :

• $\sec(50) \sin(40) + \cos(40) \csc(50) = 2$

Proof :

L.H.S =

$\sec(50) \sin(40) + \cos(40) \csc(50)$

But,

We know that,

$\bold{\sec( \alpha ) = \frac{1}{ \cos( \alpha ) } }$

And

$\bold{\csc( \alpha ) = \frac{1}{ \sin( \alpha ) } }$

Therefore,

We get,

$= \frac{ \sin(40) }{ \cos(50) } + \frac{ \cos(40) }{ \sin(50) }$

But,

We know that,

$\bold{\cos( \alpha ) = \sin(90 - \alpha )}$

And,

$\bold{\sin( \alpha ) = \cos(90 - \alpha ) }$

Therefore,

We get,

$= \frac{ \sin(40) }{ \sin(90 - 50) } + \frac{ \cos(40) }{ \cos(90 - 50) } \\ \\ = \frac{ \sin(40) }{ \sin(40) } + \frac{ \cos(40) }{ \cos(40) } \\ \\ = 1 + 1 \\ \\ = 2$

= R.H.S

Hence,

• L.H.S = R.H.S

Thus, Proved