JEE Advanced 2020

LEVELJEE Advanced

Animated Solution for Mathematics - Trigonometry: Let $x, y$ and $z$ be positive real numbers. Suppose $x, y$ and $z$ are lengths of the sides of a triangle opposite to its angles $X, Y$ and $Z$ , respectively. If $tan \frac{X}{2} + tan \frac{Z}{2} = \frac{2 y}{x + y + z}$ , then which of the following statements is/are TRUE?

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: The Spark. When you see an equation involving

tan \frac{X}{2}

and

tan \frac{Z}{2}

alongside side lengths

x, y, z

, your first instinct should be to bridge the gap between angles and sides. The half-angle formula

tan \frac{A}{2} = \frac{Δ}{s ( s - a )}

is your most powerful tool here. Visualize the triangle not just as a shape, but as a system defined by its area

Δ

and semi-perimeter

s

Hint 2

Hint 2: The Arsenal. Once you substitute the half-angle formulas into the given equation, you will find a common factor of

\frac{Δ}{s}

. Factor this out immediately. The remaining algebraic expression involving

s, x, y,

and

z

will simplify beautifully if you remember that

2 s = x + y + z

Hint 3

Hint 3: The Execution. After simplifying the expression, you will arrive at

Δ = (s - x) (s - z)

. Now, square both sides and compare this with Heron's formula

Δ^{2} = s (s - x) (s - y) (s - z)

. This comparison will lead you directly to the relationship between the sides, revealing the geometric nature of the triangle.

Visualized Solution

The Triangle Setup

Given: $tan \frac{X}{2} + tan \frac{Z}{2} = \frac{2 y}{x + y + z}$
$x, y, z$ are sides opposite to angles $X, Y, Z$ .

Half-Angle Formula

Recall the half-angle formula:
$tan \frac{A}{2} = \frac{Δ}{s ( s - a )}$
Where Area $= Δ$ and Semi-perimeter $s = \frac{x + y + z}{2}$

Substituting the Formula

Substitute into the given equation:
$\frac{Δ}{s ( s - x )} + \frac{Δ}{s ( s - z )} = \frac{2 y}{2 s}$
Notice that $x + y + z = 2 s$ .

Factoring Out Common Terms

Factor out $\frac{Δ}{s}$ :
$\frac{Δ}{s} (\frac{1}{s - x} + \frac{1}{s - z}) = \frac{y}{s}$

Taking the LCM

Combine the fractions inside the bracket:
$\frac{Δ}{s} (\frac{( s - z ) + ( s - x )}{( s - x ) ( s - z )}) = \frac{y}{s}$

Simplifying the Numerator

Simplify the numerator:
$s + s - z - x = 2 s - (x + z)$
Since $2 s = x + y + z$ :
$2 s - (x + z) = y$

Canceling Terms

Replace the numerator with $y$ :
$\frac{Δ}{s} (\frac{y}{( s - x ) ( s - z )}) = \frac{y}{s}$
Cancel $\frac{y}{s}$ from both sides:
$Δ = (s - x) (s - z)$

Squaring & Heron's Formula

Square both sides:
$Δ^{2} = (s - x)^{2} (s - z)^{2}$
Recall Heron's Formula:
$Δ^{2} = s (s - x) (s - y) (s - z)$

Equating the Areas

Equate the two expressions for $Δ^{2}$ :
$s (s - x) (s - y) (s - z) = (s - x)^{2} (s - z)^{2}$
Cancel $(s - x) (s - z)$ from both sides:
$s (s - y) = (s - x) (s - z)$

Expanding Semi-perimeter

Substitute $s = \frac{x + y + z}{2}$ :
$(\frac{x + y + z}{2}) (\frac{x - y + z}{2}) = (\frac{y + z - x}{2}) (\frac{x + y - z}{2})$

Strategic Grouping & Difference of Squares

Group terms:
$((x + z) + y) ((x + z) - y) = (y - (x - z)) (y + (x - z))$
Apply $(a + b) (a - b) = a^{2} - b^{2}$ :
$(x + z)^{2} - y^{2} = y^{2} - (x - z)^{2}$

Revealing Pythagoras

Rearrange the terms:
$(x + z)^{2} + (x - z)^{2} = 2 y^{2}$
Expand the squares:
$2 (x^{2} + z^{2}) = 2 y^{2}$
$x^{2} + z^{2} = y^{2}$

The Right-Angled Triangle

$x^{2} + z^{2} = y^{2}$ means $△ X Y Z$ is right-angled at $Y$ .
$∠ Y = \frac{π}{2}$
Since $X + Y + Z = π \Rightarrow X + Z = \frac{π}{2}$
Therefore, $Y = X + Z$

Evaluating $tan \frac{X}{2}$

Now check $tan \frac{X}{2} = \frac{Δ}{s ( s - x )}$
For our right triangle at $Y$ :
Area $Δ = \frac{1}{2} x z$
$s (s - x) = \frac{( y + z ) ^{2} - x ^{2}}{4}$

Final Conclusion

Expand denominator: $y^{2} + z^{2} + 2 y z - x^{2}$
Use $y^{2} - x^{2} = z^{2}$ :
Denominator $= 2 z^{2} + 2 y z = 2 z (y + z)$
$tan \frac{X}{2} = \frac{\frac{1}{2} x z}{\frac{2 z ( y + z )}{4}} = \frac{x}{y + z}$

00:00 / 00:00

The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Setup

Imagine you are standing before a triangle with sides

x, y, z

and angles

X, Y, Z

. You are given a seemingly complex trigonometric equation:

tan \frac{X}{2} + tan \frac{Z}{2} = \frac{2 y}{x + y + z}

In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

The Bridge of Half-Angles

To solve this, we need a bridge between the world of angles and the world of side lengths. That bridge is the half-angle formula:

tan \frac{A}{2} = \frac{( s - b ) ( s - c )}{s ( s - a )}

However, a more direct approach uses the identity

tan \frac{A}{2} = \frac{Δ}{s ( s - a )}

, where

Δ

is the area and

s

is the semi-perimeter, defined as

s = \frac{x + y + z}{2}

By substituting this into our given equation, we transform the trigonometric problem into an algebraic one:

\frac{Δ}{s ( s - x )} + \frac{Δ}{s ( s - z )} = \frac{2 y}{2 s}

Notice how the

2 s

on the right side is just

x + y + z

. This is the first sign that we are on the right track.

The Algebraic Collapse

Now, look at the left side. We have a common factor of

\frac{Δ}{s}

. Let us factor it out:

\frac{Δ}{s} (\frac{1}{s - x} + \frac{1}{s - z}) = \frac{y}{s}

The

s

in the denominator on both sides cancels out, and we are left with:

Δ (\frac{( s - z ) + ( s - x )}{( s - x ) ( s - z )}) = y

Look at the numerator inside the bracket:

s + s - x - z

. Since

2 s = x + y + z

, this numerator is simply

y

. The equation collapses to:

Δ (\frac{y}{( s - x ) ( s - z )}) = y

Canceling

y

from both sides, we get the beautiful, simple result:

Δ = (s - x) (s - z)

The Pythagorean Reveal

We are almost there. To connect this to the sides, we square both sides:

Δ^{2} = (s - x)^{2} (s - z)^{2}

Now, recall Heron's formula:

Δ^{2} = s (s - x) (s - y) (s - z)

Equating these two expressions for

Δ^{2}

, we get:

s (s - x) (s - y) (s - z) = (s - x)^{2} (s - z)^{2}

Canceling the common terms

(s - x) (s - z)

, we are left with:

s (s - y) = (s - x) (s - z)

Substituting

s = \frac{x + y + z}{2}

and simplifying, we use the difference of squares identity to arrive at:

x^{2} + z^{2} = y^{2}

This is the Pythagorean theorem! Our triangle is right-angled at

Y

. This means

∠ Y = \frac{π}{2}

, and consequently,

X + Z = \frac{π}{2}

Common Traps & FAQs

Wait, why do we use the half-angle formula instead of the Sine Rule?

Great question! While the Sine Rule is excellent for problems involving ratios of sides and sines of angles, this problem specifically gives us an expression involving

tan \frac{X}{2}

. The half-angle formula

tan \frac{A}{2} = \frac{Δ}{s ( s - a )}

is specifically designed to convert these trigonometric functions directly into algebraic expressions of sides, area, and semi-perimeter. It is the most direct path to the solution.

Why did we square both sides in the middle of the derivation?

We squared both sides to create a form that matches Heron's formula. We had

Δ = (s - x) (s - z)

, and we knew that Heron's formula involves

Δ^{2} = s (s - x) (s - y) (s - z)

. By squaring our derived equation, we obtained

Δ^{2} = (s - x)^{2} (s - z)^{2}

, which allowed us to equate the two expressions for

Δ^{2}

and cancel out common terms, leading us to the final geometric result.

Is it always a right-angled triangle?

Yes, in this specific case, the algebra forces the relationship

x^{2} + z^{2} = y^{2}

. This is the classic Pythagorean theorem, which is the necessary and sufficient condition for a triangle to be right-angled at the vertex opposite to side

y

, which is vertex

Y

. Therefore, the triangle must be right-angled at

Y

Select Answer:

Visualized Solution

The Triangle Setup

Half-Angle Formula

Substituting the Formula

Factoring Out Common Terms

Taking the LCM

Simplifying the Numerator

Canceling Terms

Squaring & Heron's Formula

Equating the Areas

Expanding Semi-perimeter

Strategic Grouping & Difference of Squares

Revealing Pythagoras

The Right-Angled Triangle

Evaluating $tan \frac{X}{2}$

Final Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Bridge of Half-Angles

The Algebraic Collapse

The Pythagorean Reveal

Similar Questions

Consider a triangle $P QR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$ , respectively. Then which of the following statements is (are) TRUE?

In a triangle $P QR, ∠ R = π /2$ . If $tan (P /2)$ and $tan (Q /2)$ are the roots of the equation $a x^{2} + b x + c = 0 (a \neq = 0)$ then.

In a triangle $A B C$ , $2 a c sin \frac{1}{2} (A - B + C) =$

In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is :

In a triangle the sum of two sides is $x$ and the product of the same sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

Consider the following statements concerning a triangle $A B C$ : (i) The sides $a, b, c$ and area $Δ$ are rational. (ii) $a, tan \frac{B}{2}, tan \frac{C}{2}$ are rational. (iii) $a, sin A, sin B, sin C$ are rational. Prove that (i) $⟹$ (ii) $⟹$ (iii) $⟹$ (i).

In a triangle $X Y Z$ , let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$ , respectively, and $2 s = x + y + z$ . If $\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$ and area of incircle of the triangle $X Y Z$ is $\frac{8 π}{3}$ , then

If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .

Let $A B C$ be a triangle such that $∠ A C B = π /6$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a = x^{2} + x + 1, b = x^{2} - 1$ and $c = 2 x + 1$ is (are)

Let $P QR$ be a triangle of area $Δ$ with $a = 2, b = 7/2$ and $c = 5/2$ , where $a, b$ , and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 s i n P - s i n 2 P}{2 s i n P + s i n 2 P}$ equals

Select Answer:

Visualized Solution

The Triangle Setup

Half-Angle Formula

Substituting the Formula

Factoring Out Common Terms

Taking the LCM

Simplifying the Numerator

Canceling Terms

Squaring & Heron's Formula

Equating the Areas

Expanding Semi-perimeter

Strategic Grouping & Difference of Squares

Revealing Pythagoras

The Right-Angled Triangle

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2X​

Final Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Bridge of Half-Angles

The Algebraic Collapse

The Pythagorean Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider a triangle PQR having sides of lengths p,q and r opposite to the angles P,Q and R, respectively. Then which of the following statements is (are) TRUE?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle PQR,∠R=π/2. If tan(P/2) and tan(Q/2) are the roots of the equation ax2+bx+c=0(a=0) then.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle ABC, 2acsin21​(A−B+C)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in a triangle ABC, cosAcosB+sinAsinBsinC=1, Show that a:b:c=1:1:2​.

Evaluating $tan \frac{X}{2}$

Consider a triangle $P QR$ having sides of lengths $p, q$ and $r$ opposite to the angles $P, Q$ and $R$ , respectively. Then which of the following statements is (are) TRUE?

In a triangle $P QR, ∠ R = π /2$ . If $tan (P /2)$ and $tan (Q /2)$ are the roots of the equation $a x^{2} + b x + c = 0 (a \neq = 0)$ then.

In a triangle $A B C$ , $2 a c sin \frac{1}{2} (A - B + C) =$

If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .