Math, asked by honeyhoney200230, 1 year ago

plz answer to my question (8 and9)
plzzzz​

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Answers

Answered by sivaprasath
7

Answer:

Step-by-step explanation:

Given :

To prove :

i) \frac{cos9+sin9}{cos9-sin9} = cot36

Proof :

We know that,

sinA+sinB = 2sin\frac{A+B}{2}cos\frac{A-B}{2}

sinA-sinB = 2sin\frac{A-B}{2}cos\frac{A+B}{2}

LHS = \frac{cos9+sin9}{cos9-sin9}

\frac{cos(90-81)+sin9}{cos(90-81)-sin9}

\frac{sin81+sin9}{sin81-sin9}

\frac{2sin\frac{81+9}{2}cos\frac{81-9}{2}}{2sin\frac{81-9}{2}cos\frac{81+9}{2}}

\frac{sin\frac{90}{2}cos\frac{72}{2}}{2sin\frac{72}{2}cos\frac{90}{2}}

\frac{sin45cos36}{sin36cos45}

\frac{sin45}{cos45} \times \frac{cos36}{sin36}  = 1 \times cot36 = cot36 = RHS

ii) sin 50 - sin70 + sin10 = 0

Proof :

⇒ LHS = sin 50 - sin70 + sin10

(sin 50 + sin10) - sin70

2sin\frac{50+10}{2}cos\frac{50-10}{2} - sin70

2sin\frac{60}{2}cos\frac{40}{2} - sin70

2sin30 \times cos20 - sin70

2(\frac{1}{2}) \times cos(90-70) - sin70

1 \times cos(90-70) - sin70

sin70 - sin70

0 = RHS

Hence, Proved,.


honeyhoney200230: tq soooo much
Anonymous: Great!
sivaprasath: neglect that 2 in 5th step (denominator)
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