Question

$if \: \tan \: theta \: = \: \sqrt{2 \: } - 1 \: prove \: that \: \sin \: theta \: . \: cos \: theta \: \frac{1}{2 \sqrt{2} }$
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Answer 1

Answer:

(1/2√2)

Step-by-step explanation:

Given Equation is tan θ = √2 - 1.

(i)

On Squaring both sides, we get

⇒ tan²θ = (√2 - 1)²

⇒ sec² θ - 1 = 2 + 1 - 2√2

⇒ sec²θ = 2 + 1 + 1 - 2√2

⇒ sec²θ = 4 - 2√2

⇒ (1/cos²θ) = 4 - 2√2

⇒ cos²θ = (1/4 - 2√2)    

⇒ cosθ = √(1/4 - 2√2)

(ii)

tan θ = √2 - 1

⇒ (1/cot θ) = √2 - 1

⇒ cot θ = (1/√2 - 1)  

On Squaring both sides, we get

⇒ cot²θ = (1/√2 - 1)²

⇒ cosec²θ - 1 = (1/2 + 1 - 2√2)

⇒ cosec²θ = (1/3 - 2√2) + 1

⇒ cosec²θ = (1 + 3 - 2√2)/(3 - 2√2)

⇒ cosec²θ = (4 - 2√2)/(3 - 2√2)

⇒ (1/sin²θ) = (4 - 2√2)/(3 - 2√2)

⇒ sin²θ = (3 - 2√2)/(4 - 2√2)

⇒ sinθ = √(3 - 2√2)/(4 - 2√2)

LHS:

sinθ * cosθ

$=\sqrt{\frac{3-2\sqrt{2}}{4-2\sqrt{2}}}*\sqrt{\frac{1}{4-2\sqrt{2}}}$

$=\sqrt{\frac{3-2\sqrt{2}}{4-2\sqrt{2}}*\frac{1}{4-2\sqrt{2}}}$

$=\sqrt{\frac{3-2\sqrt{2}}{(4-2\sqrt{2})(4-2\sqrt{2})}}$

$=\sqrt{\frac{3-2\sqrt{2}}{(4-2\sqrt{2})^2}}$

$=\sqrt{\frac{3-2\sqrt{2}}{24-16\sqrt{2}}}$

$=\sqrt{\frac{3-2\sqrt{2}}{8(3-2\sqrt{2})}}$

$=\sqrt{\frac{1}{8}}$

$=\boxed{\frac{1}{2\sqrt{2}}}$

Hope it helps!

Answer 2

Step-by-step explanation:

The answer is explained in the attachment.