Consider the following statements: 1. (sec 2θ - 1) (1 - cosec 2θ) = 1 2. sin θ (1 + cos θ) -1 + (1 + cos θ) (sin θ) -1 = 2 cosec θ
2 only
We are given two trigonometric statements and asked to determine which one is correct. To do this, we will analyze each statement individually and simplify the expressions to see if the left-hand side (LHS) equals the right-hand side (RHS).
Let's examine the first statement:
\((\sec^2 \theta - 1)(1 - \operatorname{cosec}^2 \theta) = 1\)
We can use fundamental trigonometric identities to simplify the terms in the parentheses.
Now, substitute these simplified expressions back into the LHS of Statement 1:
LHS = \((\sec^2 \theta - 1)(1 - \operatorname{cosec}^2 \theta)\)
LHS = \((\tan^2 \theta)(-\cot^2 \theta)\)
We also know that \(\cot \theta = \frac{1}{\tan \theta}\), so \(\cot^2 \theta = \frac{1}{\tan^2 \theta}\).
Substitute this into the LHS expression:
LHS = \((\tan^2 \theta)\left(-\frac{1}{\tan^2 \theta}\right)\)
LHS = \(-1\)
The RHS of Statement 1 is \(1\).
Since \(-1 \neq 1\), Statement 1 is incorrect.
Let's examine the second statement. We can rewrite the terms with negative exponents as fractions:
\(\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \operatorname{cosec} \theta\)
To simplify the LHS, find a common denominator, which is \((1 + \cos \theta)\sin \theta\):
LHS = \(\frac{\sin \theta \cdot \sin \theta}{(1 + \cos \theta)\sin \theta} + \frac{(1 + \cos \theta)(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)}\)
LHS = \(\frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta)\sin \theta}\)
Expand the term \((1 + \cos \theta)^2\) in the numerator:
\((1 + \cos \theta)^2 = 1^2 + 2(1)(\cos \theta) + \cos^2 \theta = 1 + 2 \cos \theta + \cos^2 \theta\)
Substitute this back into the numerator:
Numerator = \(\sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta\)
Rearrange the terms and use the identity \(\sin^2 \theta + \cos^2 \theta = 1\):
Numerator = \((\sin^2 \theta + \cos^2 \theta) + 1 + 2 \cos \theta\)
Numerator = \(1 + 1 + 2 \cos \theta\)
Numerator = \(2 + 2 \cos \theta\)
Numerator = \(2(1 + \cos \theta)\)
Now substitute the simplified numerator back into the LHS expression:
LHS = \(\frac{2(1 + \cos \theta)}{(1 + \cos \theta)\sin \theta}\)
Assuming \(1 + \cos \theta \neq 0\), we can cancel the term \((1 + \cos \theta)\) from the numerator and denominator:
LHS = \(\frac{2}{\sin \theta}\)
Recall that \(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\).
LHS = \(2 \cdot \frac{1}{\sin \theta}\)
LHS = \(2 \operatorname{cosec} \theta\)
The RHS of Statement 2 is \(2 \operatorname{cosec} \theta\).
Since LHS = RHS, Statement 2 is correct.
Based on our analysis:
Therefore, only Statement 2 is correct.
| Statement | Original Expression | Simplified LHS | RHS | Correctness |
|---|---|---|---|---|
| 1 | \((\sec^2 \theta - 1)(1 - \operatorname{cosec}^2 \theta) = 1\) | \(-1\) | \(1\) | Incorrect |
| 2 | \(\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \operatorname{cosec} \theta\) | \(2 \operatorname{cosec} \theta\) | \(2 \operatorname{cosec} \theta\) | Correct |
| Identity Type | Identity |
|---|---|
| Reciprocal Identities | \(\sin \theta = \frac{1}{\operatorname{cosec} \theta}, \cos \theta = \frac{1}{\sec \theta}, \tan \theta = \frac{1}{\cot \theta}\) |
| Quotient Identities | \(\tan \theta = \frac{\sin \theta}{\cos \theta}, \cot \theta = \frac{\cos \theta}{\sin \theta}\) |
| Pythagorean Identities | \(\sin^2 \theta + \cos^2 \theta = 1\) |
| \(\tan^2 \theta + 1 = \sec^2 \theta \implies \sec^2 \theta - 1 = \tan^2 \theta\) | |
| \(\cot^2 \theta + 1 = \operatorname{cosec}^2 \theta \implies \operatorname{cosec}^2 \theta - 1 = \cot^2 \theta\) |
When asked to verify a trigonometric identity, here are some general steps you can follow:
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