Math, asked by rahulkumar24032008, 6 months ago

( cos 9° + sin 9° ) / ( cos 9º – sin 9° ) = ?

Options :-

( A ) sin 54°
( B ) cos 54°
( C ) tan 54°
( D ) cot54°
​​

Answers

Answered by Anonymous
4

\bigstar Explanation \bigstar

\leadsto Solution:-

We can write 9 in terms of 45 and 54

54 - 45 = 9

Tan on both sides

tan(54 - 45) = tan(9)

We know that \rm \tan(A-B) = \frac{tanA - tanB}{1 + tanAtanB}

Therefore,

\dfrac{\tan54 - \tan45}{1+tan45tan54} = \dfrac{\sin9}{cos9}

We know that tan 45 = 1

Therefore,

\rm \dfrac{tan54 - 1}{1+tan54} = \dfrac{sin9}{cos9}

\rm \dfrac{1+tan54}{tan54-1} = \dfrac{cos9}{sin9}

Using componendo - dividendo

\dfrac{1+tan54+tan54-1}{1+tan54-tan54+1} = \dfrac{cos9+sin9}{cos9-sin9}\\\\ \dfrac{2tan54}{2} = \dfrac{cos9+sin9}{cos9-sin9}\\\\

Therefore,

\dfrac{cos9+sin9}{cos9-sin9} = tan54

\boxed{\rm Answer\:is\:option\:C}

\leadsto Concept of componendo - dividendo:-

If \frac{a}{b} = \frac{x}{y},

then \frac{(a+b)}{(a-b)} = \frac{(x+y)}{(x-y)}

Answered by parasharpraveen244
7

Answer:

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Step-by-step explanation:

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