Question

the value of tan100°+4sin100° is equal to​

Answer 1

Hola!!

A long way to go !!!! Will drive in simple way xD

First things first

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To find : tan 100 + 4 sin 100 (Formalities :p [Just Note start point])

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✓✓✓We know, $tan\, x = \frac{Sin\, x}{cos\, x}$

= $\frac{sin\, 100}{cos\, 100}+4 sin\, 100$

= $\frac{sin\, 100+4sin\, 100\: cos\, 100}{cos\, 100}$

✓✓✓ Use $Sin\: 2A = 2\: sin\, A\: cos\, A$

= $\frac{sin\, 100+2(2\: sin\, 100\: cos\, 100)}{cos\, 100}$

= $\frac{sin\, 100+2sin\, 2(100)}{cos\, 100}$

= $\frac{sin\, 100+2sin\, 2(100)}{cos\, 100}$

= $\frac{sin\, 100+2sin\, 200}{cos\, 100}$

✓✓✓ Write 2 sin 200 as sin200+ sin 200

= $\frac{(sin\, 100+sin\, 200)+ sin\, 200}{cos\, 100}$

✓✓✓ Use $Sin\, A+ Sin\, B= 2 sin\, (\frac{A+B}{2})\: cos\, (\frac{A-B}{2})$

Replace A= 200, B = 100;

$\frac{A+B}{2} =150\: \frac{A-B}{2}=50$

= $\frac{2sin\, (150)\: cos\, (50)+ sin\, 200}{cos\, 100}$ ............@Eq. 1

✓✓✓ Find Value of Sin 150, Sin 200 and Cos 100

Sin 150 = Sin (180-30) = Sin 30 = 1/2 (II Quad. sin is +Ve)

Sin 200 = Sin (270-70) = - Cos 70

cos 100 = cos (90+10) = Sin 10

Replace in @Eq. 1

= $\frac{2\, (\frac{1}{2})\: cos\, (50)+ (-cos\, 70}{-sin\, 10}$

= $\frac{cos\, 50-cos\ 70}{-sin\, 10}$

#### Ufff! looks familiar. It's cos A - Cos B.

Wait ! wait !

A should have a high value. so it is cos 70 - cos 50.

Use the (-1) from (-sin10) in the denominator for $Change\, of\, signs$

= $\frac{cos\, 70-cos\ 50}{sin\, 10}$

✓✓✓ Use $cos\, A- cos\, B= 2 sin\, (\frac{A+B}{2})\: sin\, (\frac{B-A}{2})$

Replace A = 70 , B = 50

= $\frac{A+B}{2}= 60; \frac{B-A}{2}= -10$

= $\frac{2\: sin\, 60(-sin\, 10)}{sin\, 10}$

= $-2sin\, 60$

✓✓✓ We know Sin60 = √3/2

$-2\times\frac{\sqrt{3}}{2}= -\sqrt{3}$

Look ! what we got !!! Yeah !! Your Answer

Meowwww xD

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