Question

ans my quess tiio pls pls​

Answer 1

SOLUTION

$</p><p>\qquad \quad \frac{\cos^{2} \left(45 + \theta \right) + \cos^{2} \left( 45 - \theta \right)}{\tan \left(60 +\theta \right) \times \tan \left(30 - \theta \right) } + \left( \cot 30^{\circ} + \sin 90^{\circ} \right) \times \left( \tan 60^{\circ} - \sec 0^{\circ} \right)</p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \frac{ \cos^{2} \left( 45 + \theta \right) + \sin^{2} \left( 90 - 45 + \theta \right) }{\tan \left( 60 + \theta \right) \times \cot \left( 90 - 30 + \theta \right) } + \left( \cot 30^{\circ} + \sin 90^{\circ} \right) \times \left( \tan 60^{\circ} - \sec 0^{\circ} \right)</p><p></p><p>\qquad \left( \because \cos \left( 90 - \theta \right) = \sin \theta \quad , \; \tan \left( 90 - \theta \right) = \cot \theta \right)</p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \frac{\cos^{2} \left( 45 + \theta \right) + \sin^{2} \left( 45 + \theta \right) }{ \tan \left( 60 + \theta \right) \times \cot \left( 60 + \theta \right) } + \left( \sqrt{3} + 1 \right) \cdot \left( \sqrt{3} - 1 \right) </p><p></p><p>\qquad \left( \because \cot 30^{\circ} = \sqrt{3} \quad , \; \sin 90^{\circ} , \sec 0^{\circ} = 1 \quad , \; \tan 60^{\circ} = \sqrt{3} \right) </p><p></p><p>\\ \\ </p><p></p><p>\rightarrow \quad \frac{1}{\cancel{\tan \left( 60 + \theta \right) } \times \frac{1}{\cancel{\tan \left( 60 + \theta \right) }}} + \left( \sqrt{3} + 1 \right) \cdot \left( \sqrt{3} - 1 \right)</p><p></p><p>\qquad \left( \because \sin^{2} \theta + \cos^{2} \theta = 1 \quad , \; \tan \theta = \frac{1}{\cot \theta } \right)</p><p></p><p>\\ \\ </p><p></p><p>\rightarrow \quad 1 + \left( \sqrt{3} \right)^{2} - 1^{2} \qquad \left( \because \left( a + b \right) \left( a - b \right) = a^{2} - b^{2} \right) </p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad 1 + 3 - 1</p><p></p><p>\\ \\</p><p></p><p>\rightarrow \quad \bold{3}</p><p>$

Answer 2

Answer:

$\large \bold\red{ 3}$

Step-by-step explanation:

We have to evaluate,

$\large\bold{\frac{ { \cos}^{2}(45 + \theta) + { \cos }^{2}(45 - \theta ) }{ \tan(60 + \theta ) \times \tan(30 - \theta ) } + ( \cot30 + \sin90) \times ( \tan60 - \sec0)}$

But,

We know that,

• $\bold{ \sin(90 - \alpha ) = \cos( \alpha )}$
• $\bold{ \tan(90 - \alpha ) = \cot( \alpha ) }$

Therefore,

Further simplifying,

We get,

$\large{ = \frac{ { \cos}^{2} (45 + \theta) + { \sin}^{2} (90 - 45 + \theta) }{ \tan(60 + \theta ) \times \cot(90 - 30 + \theta ) } + ( \cot30 + \sin90) \times ( \tan60 - \sec0)}$

$\large{ = \frac{ { \cos }^{2}(45 + \theta) + { \sin}^{2} (45 + \theta) }{ \tan(60 + \theta ) \times \cot(60 + \theta ) } + ( \cot30 + \sin90) \times( \tan60 - \sec0 )}$

But,

We know that,

• $\bold{{ \sin }^{2} \alpha + { \cos }^{2} \alpha = 1}$

• $\bold{ \tan( \alpha ) \times \cot( \alpha ) = 1}$

• $\bold{ \cot(30) = \sqrt{3} }$

• $\bold{ \sin(90) = 1}$

• $\bold{ \tan(60) = \sqrt{3}}$

• $\bold{ \sec(0) = 1}$

Therefore,

Putting the values,

We get,

$\large{= \frac{1}{1} + ( \sqrt{3} + 1)( \sqrt{3} - 1)}$

But,

• $\bold{(x + y)(x - y) = {x}^{2} - {y}^{2} }$

Therefore,

We get,

$= 1 + {( \sqrt{3} )}^{2} - {(1)}^{2} \\ \\ = 1 + 3 - 1 \\ \\ = \large \bold{ 3}$