Question

sin 4a + sin 5a + sin 6a/ cos 4a + cos 5a + cos 6a= tan5a

Answer 1

check the answer
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Answer 2

Answer:

as given in the question

$\frac{sin4A+sin5A+sin6A}{cos4A+cos5A+cos6A}$

has to be equal to tan5A

to prove that

by using the formula

sin C + sin D = 2 sin( $\frac{C+D}{2}$ ) .  $cos \ (\frac{C-D}{2})$

cos C + cos D = 2 cos  $(\frac{C+D}{2} )$ . $cos \ (\frac{C-D}{2} )$

there fore y using the formula

=> $\frac{2\ sin (\frac{6A+4A}{2})\times cos (\frac{6A-4A}{2} ) + sin 5A }{2\ cos (\frac{6A+4A}{2})\times cos (\frac{6A-4A}{2} ) + cos 5A}$

Taking sin 6A as C and 4A as D so that the value is perfectly divided

=> $\frac{2\ sin {5A}\times cos {A} + sin 5A }{2\ cos 5A\times cos A + cos 5A}$

Now , take common sin 5A in the numerator and cos 5A in the denominator

=> $\frac{ sin {5A}( 2cos {A} + 1) }{ cos 5A(2 cos A + 1)}$

by cancelling ( 2 cos A +1) in the numerator and the denominator

gives us

=> $\frac{sin5A}{cos 5A}$

=> $tan5A$

hence proved

To learn more about Trigonometric equation :

https://brainly.in/question/6820230

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