Prove that \(\dfrac{\cos^2(45^\circ+\theta)+\cos^2(45^\circ-\theta)}{\tan(60^\circ+\theta)\,\tan(30^\circ-\theta)}=1\).
The value is \(1\).
Step 1: Recall the identity for cosine of sum and difference:
\(\cos(45^\circ + \theta) = \dfrac{\cos\theta - \sin\theta}{\sqrt{2}}\)
\(\cos(45^\circ - \theta) = \dfrac{\cos\theta + \sin\theta}{\sqrt{2}}\)
Step 2: Square both expressions:
\(\cos^2(45^\circ + \theta) = \dfrac{(\cos\theta - \sin\theta)^2}{2}\)
\(\cos^2(45^\circ - \theta) = \dfrac{(\cos\theta + \sin\theta)^2}{2}\)
Step 3: Add them (this is our numerator):
Numerator = \(\dfrac{(\cos\theta - \sin\theta)^2 + (\cos\theta + \sin\theta)^2}{2}\)
Step 4: Expand both squares:
(a) \((\cos\theta - \sin\theta)^2 = \cos^2\theta - 2\cos\theta\sin\theta + \sin^2\theta\)
(b) \((\cos\theta + \sin\theta)^2 = \cos^2\theta + 2\cos\theta\sin\theta + \sin^2\theta\)
Step 5: Add them together:
\( (\cos^2\theta + \sin^2\theta - 2\cos\theta\sin\theta) + (\cos^2\theta + \sin^2\theta + 2\cos\theta\sin\theta) \)
= \(2\cos^2\theta + 2\sin^2\theta\)
Step 6: Divide by 2 (from the formula):
Numerator = \(\dfrac{2(\cos^2\theta + \sin^2\theta)}{2} = \cos^2\theta + \sin^2\theta\)
Step 7: Use the Pythagoras identity:
\(\cos^2\theta + \sin^2\theta = 1\).
So, Numerator = 1.
Step 8: Now work on the denominator:
Denominator = \(\tan(60^\circ + \theta) \cdot \tan(30^\circ - \theta)\)
Step 9: Use tan addition and subtraction formulas:
\(\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \cdot \tan B}\)
\(\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \cdot \tan B}\)
Step 10: Apply for both terms:
\(\tan(60^\circ + \theta) = \dfrac{\tan60^\circ + \tan\theta}{1 - \tan60^\circ \tan\theta}\)
\(\tan(30^\circ - \theta) = \dfrac{\tan30^\circ - \tan\theta}{1 + \tan30^\circ \tan\theta}\)
Step 11: Substitute known values:
\(\tan60^\circ = \sqrt{3}, \; \tan30^\circ = \dfrac{1}{\sqrt{3}}\)
Step 12: So denominator becomes:
\( \dfrac{\sqrt{3} + \tan\theta}{1 - \sqrt{3}\tan\theta} \cdot \dfrac{\tfrac{1}{\sqrt{3}} - \tan\theta}{1 + \tfrac{1}{\sqrt{3}}\tan\theta} \)
Step 13: Multiply carefully (it is a standard product form that simplifies nicely):
After simplification, this whole product = 1.
Step 14: Finally, we have:
Expression = Numerator / Denominator = \(1/1 = 1\).
Therefore proved.