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If $a 2 − b 2 sin θ + 2 a b cos θ = a 2 + b 2,$ then what is the value of $tan θ ?$

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a 2 − b 2 2 a b

a 2 + b 2 2 a b

a 2 − b 2 2 a

a 2 + b 2 2 b

answer is A.

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Detailed Solution

Given that,

a 2 − b 2 s i n θ + 2 a b cos θ = a 2 + b 2

.
Dividing both sides by

cos θ

, we get,

a 2 − b 2 s i n θ + 2 a b cos θ = a 2 + b 2

⇒ a 2 − b 2 s i n θ + 2 a b cos θ cos θ = a 2 + b 2 cos θ

Splitting terms, we get,

⇒ a 2 − b 2 s i n θ cos θ + 2 a b cos θ cos θ = a 2 + b 2 cos θ

⇒ a 2 − b 2 tan θ + 2 a b = a 2 + b 2 sec θ

…(1)

Since tan θ = sin θ cos θ, sec θ = 1 cos θ

Squaring both sides in (1), we get,

⇒ a 2 − b 2 tan θ + 2 a b 2 = a 2 + b 2 sec θ 2

⇒ a 2 − b 2 2 tan 2 θ + 4 a 2 b 2 + 4 a b a 2 − b 2 tan θ = a 2 + b 2 2 sec 2 θ [(a + b) 2 = a 2 + 2 a b + b 2]

⇒ a 2 − b 2 2 tan 2 θ − a 2 + b 2 2 sec 2 θ + 4 a 2 b 2 + 4 a b a 2 − b 2 tan θ = 0

Applying the identity from

1 + tan 2 θ = sec 2 θ

, we get,

⇒ a 2 − b 2 2 tan 2 θ − a 2 + b 2 2 (1 + tan 2 θ) + 4 a 2 b 2 + 4 a b a 2 − b 2 tan θ = 0

⇒ a 2 − b 2 2 tan 2 θ − a 2 + b 2 2 − a 2 + b 2 2 tan 2 θ + 4 a 2 b 2 + 4 a b a 2 − b 2 tan θ = 0

Using the first and third terms together, we get,

tan 2 θ a 2 − b 2 2 − a 2 + b 2 2 − a 2 + b 2 2 + 4 a 2 b 2 + 4 a b a 2 − b 2 tan θ = 0

Expanding the first term and canceling terms, we get,

− 4 a 2 b 2 tan 2 θ − a 2 + b 2 2 + 4 a 2 b 2 + 4 a b a 2 − b 2 tan θ = 0

Opening whole square terms and simplifying, we get, Question Image

∵ a 2 + b 2 − 2 a b = (a − b) 2

gives

4 a 2 b 2 tan 2 θ − 4 a b a 2 − b 2 tan θ + a 2 − b 2 2 = 0

…(2)
Solving (2) by quadratic formula to find the value of

tan θ

with the help of

a x 2 + b x + c = 0

then

x 1, 2 = − b ± b 2 − 4 a c 2 a

, we get,
Comparing

4 a 2 b 2 tan 2 θ − 4 a b a 2 − b 2 tan θ + a 2 − b 2 2 = 0

with standard quadratic equation, we get,

a = 4 a 2 b 2 b = − 4 a b a 2 − b 2 c = a 2 − b 2 2

Applying the quadratic formula, we get,

tan θ 1, 2 = 4 a b a 2 − b 2 ± 4 a b a 2 − b 2 2 − 4 4 a 2 b 2 a 2 − b 2 2 8 a 2 b 2

tan θ 1, 2 = 4 a b a 2 − b 2 ± 16 a 2 b 2 a 2 − b 2 2 − 16 a 2 b 2 a 2 − b 2 2 8 a 2 b 2

Canceling the term under the square root, we get,

tan θ 1, 2 = 4 a b a 2 − b 2 ± 0 8 a 2 b 2

⇒ tan θ = 4 a b a 2 − b 2 8 a 2 b 2

⇒ tan θ = 4 a b a 2 − b 2 2 a b 8 a 2 b 2

⇒ tan θ = a 2 − b 2 2 a b

Thus,

tan θ = a 2 − b 2 2 a b .

Hence, option 1 is correct.

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