TBQ - Prove the following III — task. Maths TNSB Mentoring, Class 10 Crash course.

(i) If

\sin y + \cos y = \sqrt{3}

, then prove that

\tan y + \cot y = 1

Proof:

Consider

\sin y + \cos y = \sqrt{3}

Squaring on both sides, we get:

----- (1)

Now consider LHS.

LHS \(=\)

\tan y + \cot y

\(=\)

1

\(=\) RHS

Hence, we proved.

(ii) If

\sqrt{3} \cos y - \sin y = 0

, then verify that \(tan \ 3 \theta = \frac{3 \ tan \ \theta - tan^3 \ \theta}{1 - 3 \ \tan^2 \ \theta}\)

Proof:

Consider

\sqrt{3} \cos y - \sin y = 0

Now, consider LHS.

LHS \(= tan \ 3 \theta\)

\(=\)

---- (3)

Now consider RHS.

RHS \(= \frac{3 \ tan \ \theta - tan^3 \ \theta}{1 - 3 \ \tan^2 \ \theta}\)

\(=\)

---- (4)

From equations (\(3\)) and (\(4\)), we can see that LHS \(=\) RHS.

Therefore, \(tan \ 3 \theta = \frac{3 \ tan \ \theta - tan^3 \ \theta}{1 - 3 \ \tan^2 \ \theta}\)

Hence, we proved.

Answer variants:

\frac{1}{1}

\tan 180 °

\(1\)

\cot y = \cot 60 °

\tan 3 (60 °)

\frac{\cos y}{\sin y} = \frac{1}{\sqrt{3}}

\sin y \cos y = 1

1 + 2 \sin y \cos y = 3

\frac{\sin y}{\cos y} + \frac{\cos y}{\sin y}

0

\sqrt{3} \cos y = \sin y

y = 60 °

\frac{\sin^{2} y + \cos^{2} y}{\sin y \cos y}

\cot y = \frac{1}{\sqrt{3}}

{(\sin y + \cos y)}^{2} = {(\sqrt{3})}^{2}

\sin^{2} y + \cos^{2} y + 2 \sin y \cos y = 3

2 \sin y \cos y = 2

Did you find an error?

8. TBQ - Prove the following III