What is the value of (1 + cot A + tan A)(sin A - cos A) \(\rm \left(\frac{\sin A\cos A}{\sin^3A-\cos^3A}\right) \)?
1
The question asks for the value of a given trigonometric expression involving $\sin A$, $\cos A$, $\tan A$, and $\cot A$. To find the value, we need to simplify the expression using basic trigonometric identities and algebraic formulas.
The given expression is:
\((1 + \cot A + \tan A)(\sin A - \cos A) \left(\frac{\sin A\cos A}{\sin^3A-\cos^3A}\right)\)
We can express $\cot A$ and $\tan A$ in terms of $\sin A$ and $\cos A$:
\(\cot A = \frac{\cos A}{\sin A}\)
\(\tan A = \frac{\sin A}{\cos A}\)
Substitute these into the first factor:
\(1 + \cot A + \tan A = 1 + \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}\)
Find a common denominator, which is $\sin A \cos A$:
\(= \frac{\sin A \cos A}{\sin A \cos A} + \frac{\cos A \cdot \cos A}{\sin A \cos A} + \frac{\sin A \cdot \sin A}{\sin A \cos A}\)
\(= \frac{\sin A \cos A + \cos^2 A + \sin^2 A}{\sin A \cos A}\)
Using the fundamental trigonometric identity $\sin^2 A + \cos^2 A = 1$:
\(= \frac{\sin A \cos A + 1}{\sin A \cos A}\)
This is a difference of cubes, which can be factored using the algebraic identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Here, \(a = \sin A\) and \(b = \cos A\).
\(\sin^3 A - \cos^3 A = (\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)\)
Again using the identity $\sin^2 A + \cos^2 A = 1$:
\(= (\sin A - \cos A)(1 + \sin A \cos A)\)
Now substitute the simplified forms back into the original expression:
\(\left(\frac{1 + \sin A \cos A}{\sin A \cos A}\right) (\sin A - \cos A) \left(\frac{\sin A\cos A}{(\sin A - \cos A)(1 + \sin A \cos A)}\right)\)
Let's rewrite the entire expression as a single fraction to see the terms clearly:
\(= \frac{(1 + \sin A \cos A)}{\sin A \cos A} \times (\sin A - \cos A) \times \frac{\sin A\cos A}{(\sin A - \cos A)(1 + \sin A \cos A)}\)
\(= \frac{(1 + \sin A \cos A) \times (\sin A - \cos A) \times (\sin A \cos A)}{(\sin A \cos A) \times (\sin A - \cos A) \times (1 + \sin A \cos A)}\)
We can observe that there are common factors in the numerator and the denominator:
Assuming $\sin A \ne 0$, $\cos A \ne 0$, $\sin A \ne \cos A$, and $1 + \sin A \cos A \ne 0$ (which are generally true unless A takes specific values that would make the original expression undefined anyway), we can cancel these terms:
\(= \frac{\cancel{(1 + \sin A \cos A)} \times \cancel{(\sin A - \cos A)} \times \cancel{(\sin A \cos A)}}{\cancel{(\sin A \cos A)} \times \cancel{(\sin A - \cos A)} \times \cancel{(1 + \sin A \cos A)}}\)
After cancellation, the expression simplifies to:
\(= 1\)
The value of the given trigonometric expression \((1 + \cot A + \tan A)(\sin A - \cos A) \left(\frac{\sin A\cos A}{\sin^3A-\cos^3A}\right)\) simplifies to \(1\).
| Expression Part | Simplified Form | Identity Used |
|---|---|---|
| \(1 + \cot A + \tan A\) | \(\frac{1 + \sin A \cos A}{\sin A \cos A}\) | \(\cot A = \frac{\cos A}{\sin A}\), \(\tan A = \frac{\sin A}{\cos A}\), \(\sin^2 A + \cos^2 A = 1\) |
| \(\sin^3 A - \cos^3 A\) | \((\sin A - \cos A)(1 + \sin A \cos A)\) | \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\), \(\sin^2 A + \cos^2 A = 1\) |
| Identity | Description |
|---|---|
| \(\tan A = \frac{\sin A}{\cos A}\) | Tangent in terms of sine and cosine. |
| \(\cot A = \frac{\cos A}{\sin A}\) | Cotangent in terms of sine and cosine. |
| \(\sin^2 A + \cos^2 A = 1\) | The fundamental Pythagorean identity. |
| \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) | Difference of cubes algebraic identity. |
Simplifying trigonometric expressions is a common task in trigonometry. The general strategy involves:
It's crucial to remember the domain restrictions for functions like $\tan A$ and $\cot A$ and for expressions involving denominators to avoid division by zero.
The given equation can be reduced to
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