Which of the following will satisfy a2 = b2 + (ab)2 for the values a and b?
a = cot x, b = cos x
The problem asks us to find a pair of trigonometric values for 'a' and 'b' that satisfies the given equation: \(a^2 = b^2 + (ab)^2\).
We are provided with four options, each giving specific trigonometric expressions for 'a' and 'b'. We need to substitute these values into the equation and check which pair makes the equation true.
Let's examine each option by substituting the values of 'a' and 'b' into the equation \(a^2 = b^2 + (ab)^2\) and simplifying both sides.
Substitute \(a = \sin x\) and \(b = \cot x\) into the equation:
Left Hand Side (LHS): \(a^2 = (\sin x)^2 = \sin^2 x\)
Right Hand Side (RHS): \(b^2 + (ab)^2 = (\cot x)^2 + (\sin x \cdot \cot x)^2\)
We know that \(\cot x = \frac{\cos x}{\sin x}\). Substitute this into the RHS:
RHS: \(\cot^2 x + \left(\sin x \cdot \frac{\cos x}{\sin x}\right)^2\)
RHS: \(\cot^2 x + (\cos x)^2\)
RHS: \(\cot^2 x + \cos^2 x\)
Comparing LHS and RHS: Is \(\sin^2 x = \cot^2 x + \cos^2 x\)? This is generally false.
So, option 1 does not satisfy the equation.
Substitute \(a = \cos x\) and \(b = \tan x\) into the equation:
Left Hand Side (LHS): \(a^2 = (\cos x)^2 = \cos^2 x\)
Right Hand Side (RHS): \(b^2 + (ab)^2 = (\tan x)^2 + (\cos x \cdot \tan x)^2\)
We know that \(\tan x = \frac{\sin x}{\cos x}\). Substitute this into the RHS:
RHS: \(\tan^2 x + \left(\cos x \cdot \frac{\sin x}{\cos x}\right)^2\)
RHS: \(\tan^2 x + (\sin x)^2\)
RHS: \(\tan^2 x + \sin^2 x\)
Comparing LHS and RHS: Is \(\cos^2 x = \tan^2 x + \sin^2 x\)? This is generally false.
So, option 2 does not satisfy the equation.
Substitute \(a = \cot x\) and \(b = \cos x\) into the equation:
Left Hand Side (LHS): \(a^2 = (\cot x)^2 = \cot^2 x\)
Right Hand Side (RHS): \(b^2 + (ab)^2 = (\cos x)^2 + (\cot x \cdot \cos x)^2\)
We know that \(\cot x = \frac{\cos x}{\sin x}\). Substitute this into the RHS:
RHS: \(\cos^2 x + \left(\frac{\cos x}{\sin x} \cdot \cos x\right)^2\)
RHS: \(\cos^2 x + \left(\frac{\cos^2 x}{\sin x}\right)^2\)
RHS: \(\cos^2 x + \frac{\cos^4 x}{\sin^2 x}\)
To add these terms, find a common denominator, which is \(\sin^2 x\):
RHS: \(\frac{\cos^2 x \sin^2 x}{\sin^2 x} + \frac{\cos^4 x}{\sin^2 x}\)
RHS: \(\frac{\cos^2 x \sin^2 x + \cos^4 x}{\sin^2 x}\)
Factor out \(\cos^2 x\) from the numerator:
RHS: \(\frac{\cos^2 x (\sin^2 x + \cos^2 x)}{\sin^2 x}\)
Using the fundamental trigonometric identity \(\sin^2 x + \cos^2 x = 1\):
RHS: \(\frac{\cos^2 x (1)}{\sin^2 x}\)
RHS: \(\frac{\cos^2 x}{\sin^2 x}\)
Using the identity \(\cot x = \frac{\cos x}{\sin x}\):
RHS: \((\cot x)^2 = \cot^2 x\)
Comparing LHS and RHS: We found LHS = \(\cot^2 x\) and RHS = \(\cot^2 x\). Since LHS = RHS, this option satisfies the equation.
Substitute \(a = \sin x\) and \(b = \tan x\) into the equation:
Left Hand Side (LHS): \(a^2 = (\sin x)^2 = \sin^2 x\)
Right Hand Side (RHS): \(b^2 + (ab)^2 = (\tan x)^2 + (\sin x \cdot \tan x)^2\)
RHS: \(\tan^2 x + \sin^2 x \tan^2 x\)
Factor out \(\tan^2 x\):
RHS: \(\tan^2 x (1 + \sin^2 x)\)
Comparing LHS and RHS: Is \(\sin^2 x = \tan^2 x (1 + \sin^2 x)\)? This is generally false.
So, option 4 does not satisfy the equation.
Based on the step-by-step verification, the pair of values \(a = \cot x\) and \(b = \cos x\) is the only option among the given choices that satisfies the trigonometric equation \(a^2 = b^2 + (ab)^2\).
Here are some fundamental trigonometric identities used in solving this problem:
| Identity | Description |
|---|---|
| \(\tan x = \frac{\sin x}{\cos x}\) | Tangent in terms of Sine and Cosine |
| \(\cot x = \frac{\cos x}{\sin x}\) | Cotangent in terms of Sine and Cosine |
| \(\sin^2 x + \cos^2 x = 1\) | Pythagorean Identity |
| \(\cot^2 x + 1 = \csc^2 x\) | Pythagorean Identity (derived) |
| \(\tan^2 x + 1 = \sec^2 x\) | Pythagorean Identity (derived) |
Solving trigonometric equations often involves using identities to simplify the expression or equation. The goal is usually to express everything in terms of a single trigonometric function or to reach a known identity.
In this problem, we tested specific values, but for general trigonometric equations, you might need to:
Practicing with different identities helps in simplifying and solving complex trigonometric problems like verifying if certain values satisfy an equation.
If sec x – cos x = 4, then what will be the value of \({{(1 \ + \ cos^2 x)} \over cos x}\)?
In ΔPQR, ∠Q = 90°. If tan R = \(\frac{1}{3}\) then what is the value of \(\rm \frac{secP(cosR+sinP)}{cosecR(sinR-cosecP)}\) ?
What is the value of 5 sin 260° + 7 sin 245° + 8 cos 245° ?
If \(\rm \frac{\cos^2 θ}{\cot^2 θ + \sin^2 θ -1} = 3, 0^\circ < θ < 90^\circ\) , then the value of (tan θ + cosec θ) is:
If sec 2θ + tan 2θ = 5/3, then what is the value of tan2θ?
In ΔABC, right angled at B, if tanA = 1/2, then the value of \(\frac{sin A(cos C + cos A)}{cos C (sin C - sin A)}\) is:
The expression \(\rm \frac{(1-sin\,θ+cos\,θ)^2(1-cos\,θ)sec^3\,θ cosec^2\,θ}{(sec\,θ -tan\,θ)(tan\,θ+cot\,θ)} \) , 0° < θ < 90°, is equal to:
If \(\frac {\sec θ + \tan θ }{\sec θ - \tan θ } = 2 \frac {51}{79}, \) then the value of sin θ is equal to:
If \(\rm \frac{sec\,θ-tan\,θ}{sec\,θ+tan\,θ}=\frac{1}{7}\) , θ lies in first quadrant, then the value of \(\rm \frac{cosec\,\theta+cot^2\,\theta}{cosec\,\theta-cot^2\,\theta}\) is:
If cos θ = \(\frac{12}{13}\) , then the value of \(\rm \frac{sin\,\theta(1-tan\,\theta)}{tan\,\theta(1+cosec\,\theta)}\) is:
What is the ratio of the greatest to the smallest value of 2 – 2 sin x – sin 2x, 0 ≤ x ≤ (π/2)?
If sinθ = \(\frac{4}{5}\) , Find the value of sin3θ
If x, y are acute angles, where 0 < x + y < 90° and sin(3x - 40°) = cos (3y + 40°), then the value of tan (x + y) is equal to
What is (1 + cot θ - cosec θ)(1 + tan θ + sec θ) equal to?
If cos x = p/q and 0° < x < 90°, then the value of tan x is: