Question

In a triangle ABC C= 90° and tan A =1√3 find the value of : (¹) (sin A cos B + cos A sin B ) 2 (cos A cos B -sin sin A sin B )​

Answer 1

Answer:

0

Step-by-step explanation:

$tan A = 1/\sqrt{3}$

We know that $tanA = sinA/cosA$

Let $AC = \sqrt{3x}$ and $BC = 1x$

Applying the Pythagoras Theorem,

$AC^{2} + BC^{2} = AB^{2}$

$(\sqrt{3}x)^{2} + x^{2} = AB^{2}$

$3x^{2} + x^{2} = AB^{2}$

$4x^{2} = AB^{2}$

$2x = AB$

$sinA = 1/2\\\\cosB = sinA = 1/2\\\\cosA = \sqrt{3}/2\\\\sinB = cosA = \sqrt{3}/2$

Breaking the sum into two parts, we get

$sinA cosB + cosA sinB = (1/2)^{2} + (\sqrt{3}/2)^{2} = 1/4 + 3/4 = 4/4 = 1$

$2(cosA cosB - sinA sinB) = 2[(\sqrt{3}/2 * 1/2) - (1/2 * \sqrt{3}/2)] = 2(\sqrt{3}/4 - \sqrt{3}/4) = 0$

∴ Multiplying the above two parts, we obtain 0

Answer 2

$\overline{\underline{\boxed{\sf GIVEN \darr}}}$

In a triangle ABC C= 90° and tan A =1√3 find the value of : (¹) (sin A cos B + cos A sin B ) 2 (cos A cos B -sin sin A sin B )

$\overline{\underline{\boxed{\sf HOW \: TO \: DO \darr}}}$

Step 1 : First we will take any constant multiple

Step 2 : Then we wil put the values of tan A with the multiple

Step 3 : Then we need to apply Pythagoras theorem to find the Hypotenuse

Step 4 : Then applying the formula to get sin and cos we will put the values

$\overline{\underline{\boxed{\sf SOLUTION \darr}}}$

Given that :

tan A = 1/√3

Let the constant multiple be k

tan A = Opposite/Adjacent

According to the given picture

tan A = BC/AC

So,

• tan A = 1k/√3k

• tan A = 1/√3

Now to get the Pythagoras Theorem :

In ∆ACB , ∠C = 90°

(Hypotentuse)² = (Perpendicular)² + (Base)²

• (AB)² = (BC)² + (AC)²

• (AB)² = (1k)² + (√3k)²

• (AB)² = 1k² + 3k²

• (AB)² = 4k²

• AB = √4k²

• AB = 2k

Now, according to the condition which is given is :

1) sin A cos B + cos A sin B

So,

sin A = Perpendicular/Hypotentuse = BC/AB = 1k/2k = 1/2

cos B = Base/Hypotentuse = AC/AB = √3k/2k = √3/2

cos A = Base/Hypotentuse = BC/AB = 1k/2k = 1/2

sin B = Perpendicular/Hypotentuse = AC/AB = √3/2

Hence, putting all the values we get

• 1/2 × √3/2 + 1/2 × √3/2

• √3/4 + √3/4

• 2√3/4

• √3/2

Hence, the answer is √3/2

2) cos A cos B - sin A sin B

cos A = Base/Hypotentuse = BC/AB = 1k/2k = 1/2

cos B = Base/Hypotentuse = AC/AB = √3k/2k = √3/2

sin A = Perpendicular/Hypotentuse = BC/AB = 1k/2k = 1/2

sin B = Perpendicular/Hypotentuse = AC/AB = √3/2

So, putting the values we get

• 1/2 × √3/2 - 1/2 × √3/2

• √3/4 - √3/4

• 0

Hence, the answer is 0