Question

If the angles of $\triangle \mathrm{ABC}$ are in ratio $1: 1: 2$, respectively (the largest angle being angle C), then the value of $\frac{\sec A}{cosec B}-\frac{\tan A}{\cot B}$ is

(a) 0

(b) $1 / 2$

(c) 1

(d) $\sqrt{3} / 2$

Answer 1

Answer:

0

Step-by-step explanation:

Given that the angles of a triangle are in ratio of 1:1:2 where C is the largest angle of a triangle ABC. Inorder to solve this question, you must be aware about The Angle Sum Property Of Triangle and Values of trig. ratios.

According to the Angle Sum Property Of Triangle, the sum of all the angles of a triangle is always equal to 180°.

I'll be mentioning all the values of trig. ratios at the end.

So let's start our calculations!

$\rule{280}{1}$

Solution:-

Let's assume the ratio of angles be:

• A : B : C = x : x : 2x

By angle sum property, we know:

➝ ∠A + ∠B + ∠C = 180°

➝ x + x + 2x = 180°

➝ 4x = 180°

➝ x = 180°/4

➝ x = 45°

Therefore the angles of the triangle are:

• A = x = 45°
• B = x = 45°
• C = 2x = 90°

Now we have to find the value of:

$\sf \implies\dfrac{sec \: A}{cosec \: B}-\dfrac{tan \: A}{cot \: B}$

$\sf \implies\dfrac{sec \: 45^ \circ}{cosec \: 45 ^\circ}-\dfrac{tan \:45^ \circ }{cot \: 45^ \circ}$

$\sf \implies\dfrac{ \frac{1}{ \sqrt{2} } }{ \frac{1}{ \sqrt{2} } }-\dfrac{1 }{1}$

$\sf \implies1-1$

$\sf \implies0$

So the required answer is:

$\boxed{ \tt \implies\dfrac{sec \: A}{cosec \: B}-\dfrac{tan \: A}{cot \: B} = 0}$

Option (A) is correct.

More Information:

Values of trigonometric ratios:

sin (θ)

• 0° = 0
• 30° = 1/2
• 45° = 1/√2
• 60° = √3/2
• 90° = 1

cos (θ)

• 0° = 1
• 30° = √3/2
• 45° = 1/√2
• 60° = 1/2
• 90° = 1

tan (θ)

• 0° = 0
• 30° = 1/√3
• 45° = 1
• 60° = √3
• 90° = Undefined

Values of cosec(θ), sec(θ) and cot(θ) are reciprocal of the values of sin(θ), cos(θ) and tan(θ) respectively.