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1 year ago

Sin2A+Sin2B÷Sin2A-Sin2B=Tan(A+B)÷Tan(A-B) Prove that

Mathematics

1 answer:

igor_vitrenko [27]1 year ago

8 0

Take L.H.S sin2A+sin2B/sin2A-sin2B

= sin2A+sin2B/sin2A-sin2B

Put

[sinC+sinD = 2sin(C+D)/2cos(C-D)/2]

[sinC-sinD = 2cos(C+D)/2.sin(C-D)/2]

= 2 sin(2A+2B)/2 cos(2A-2B)/2 / 2 cos(2A+2B) sin(2A-2B)

= sin(A+B).cos(A-B)/cos(A+B).sin(A-B)

= sin(A+B)/cos(A+B) . cos(A-B)/sin(A-B)

= tan(A+B).cot(A-B)

= tan(A+B).1/tan(A-B)

= tan(A+B)/tan(A-B)

∴ Hence we proved sin2A+sin2B/sin2A-sin2B=tan(A+B)/tan(A-B)

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Solve 5r2 – 12 = 68. {±4} {±5} {±3} {±6}

sashaice [31]

Hey there!

Assuming you meant

$5r^2 - 12 = 68$

If so, follow these steps so it can be a little bit easier to solve

Firstly, we have to $add$ by $12$ on each of your sides

$5r^2 -12 +12 \\ \\ 68 + 12$

$5r^2 = 5r^2 \\ \\ 68 + 12 = 80$

We get: $5r^2 =80$

Now, we have $divide$ by $5$ on each of your sides

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We get: $r^2 = 16$

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Good luck on your assignment and enjoy your day!

~ $LoveYourselfFirst:)$

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2 years ago

Sin2A+Sin2B÷Sin2A-Sin2B=Tan(A+B)÷Tan(A-B) Prove that​

Sin2A+Sin2B÷Sin2A-Sin2B=Tan(A+B)÷Tan(A-B) Prove that