Question

If costheta - sintheta = root2sintheta, prove that costheta + sintheta=root2costheta
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Answer 1

Question :- if cosθ - sinθ = √2sinθ, prove that cosθ + sinθ = √2cosθ . ?

Solution :-

→ cosθ - sinθ = √2sinθ

→ cosθ = √2sinθ + sinθ

→ cosθ = sinθ(√2 + 1)

→ (sinθ / cosθ) = 1/(√2 + 1)

Rationalize RHS part now,

→ (sinθ / cosθ) = 1/(√2 + 1) * {(√2 - 1) / (√2 - 1)}

→ (sinθ / cosθ) = (√2 - 1)/{(√2 + 1)(√2 - 1)}

using (a + b)(a - b) = a² - b² in RHS denominator now,

→ (sinθ / cosθ) = (√2 - 1)/ (2 - 1)

→ (sinθ / cosθ) = (√2 - 1)

→ sinθ = cosθ(√2 - 1)

→ sinθ = √2cosθ - cosθ

→ sinθ + cosθ = √2cosθ . (Hence, Proved).

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Answer 2

Answer:

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