AP EAMCET · Maths · Straight Lines

For $α \in [0, \frac{π}{2}]$ , the angle between the lines represented by $[x \cos θ - y] [(\cos θ + \tan α) x - (1 - \cos θ \cdot \tan α) y] = 0$ is

A $α$
B $θ$
C $θ + α$
D $θ - α$

Verified Solution

Answer & Solution

Correct Answer

(A) $α$

Step-by-step Solution

Detailed explanation

Given, xcosθ-ycosθ+tanαx-1-cosθ·tanαy=0 Now taking line L1→xcosθ-y=0 And L2→cosθ+tanαx-1-cosθ·tanαy=0 Now slope of line L1 is m1=cosθ and slope of line L2 is m2=cosθ+tanα1-cosθ.tanα…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

If $\cos \alpha+\cos \beta=a$ and $\sin \alpha+\sin \beta=b$, then match the items given in List $-A$ with those of their values in List-B

List-A	Liat-B
(I) $\tan \left(\frac{\alpha+\beta}{2}\right)=$	(a) $\frac{ b }{ a }$
(II) $\cos (\alpha+\beta)=$	(b) $\frac{2 a b}{a^2+b^2}$
(III) $\sin (\alpha+\beta)^{-}$	(c) $\frac{2 a b}{a^2-b^2}$
(IV) $\tan (\alpha+\beta)=$	(d) $\frac{a^2-b^2}{a^2+b^2}$
	(e) $\frac{a^2+b^2}{a^2-b^2}$

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List-A	Liat-B
(I) \(\tan \left(\frac{\alpha+\beta}{2}\right)=\)	(a) \(\frac{ b }{ a }\)
(II) \(\cos (\alpha+\beta)=\)	(b) \(\frac{2 a b}{a^2+b^2}\)
(III) \(\sin (\alpha+\beta)^{-}\)	(c) \(\frac{2 a b}{a^2-b^2}\)
(IV) \(\tan (\alpha+\beta)=\)	(d) \(\frac{a^2-b^2}{a^2+b^2}\)
	(e) \(\frac{a^2+b^2}{a^2-b^2}\)

For α∈0,π2, the angle between the lines represented by xcosθ-ycosθ+tanαx-1-cosθ·tanαy=0 is

Step-by-step Solution

For $α \in [0, \frac{π}{2}]$ , the angle between the lines represented by $[x \cos θ - y] [(\cos θ + \tan α) x - (1 - \cos θ \cdot \tan α) y] = 0$ is