Question

Please help me to do this ​

Answer 1

Answer:

2

Step-by-step explanation:

 We know, cos( 90 - A ) = sinA ;

                   sec( 90 - A ) = cosecA ;

               sin^2 A + cos^2 A = 1

Here,

⇒ cos²43° + cos²47° + ( sec50°/cosec40° )

       cos²43° = cos( 90° - 47° )

       sec50° = sec( 90° - 40° )

⇒ cos²( 90 - 47 ) + cos²47° + ( sec(90-40)/cosec40 )

⇒ sin²47° + cos²47° + ( cosec40°/cosec40° )

⇒ 1 + 1

⇒ 2

Answer 2

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$\boxed{ \blue{2}}$

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Some important results

• cos(90-x) = sinx
• sec(90-x) = cosec x

the functions changes at 90•

since

cos (43) = sin(90-43) = sin(47)

sec(50)=cosec(90-50) =cosec(40)

$\rightarrow \: {( \cos43)}^{2} + {( \cos47)}^{2} + \frac{{( \sec50)}^{2} }{ { (\csc40) }^{2} }$

$\rightarrow \: {( \cos43)}^{2} + {( \sin(90 - 47) )}^{2} + \frac{{( \sec(90 - 50))}^{2} }{ { (\csc40) }^{2} }$

....... from 1 and 2

$\rightarrow \: {( \cos43)}^{2} + {( \sin(43) )}^{2} + \frac{{( \csc(40) )}^{2} }{ { (\csc40) }^{2} }$

we know

sin^2 x + cos^x = 1

$\rightarrow 1 + 1$

$\rightarrow \boxed{ \blue{2}}$