Trigonometric Identities Question 127
Question: If $ a{{\sin }^{2}}x+b{{\cos }^{2}}x=c, $ $ b,{{\sin }^{2}}y+a,{{\cos }^{2}}y=d $ and $ a,\tan x=b,\tan y, $ then $ \frac{a^{2}}{b^{2}} $ is equal to
Options:
A) $ \frac{(b-c)(d-b)}{(a-d)(c-a)} $
B) $ \frac{(a-d)(c-a)}{(b-c)(d-b)} $
C) $ \frac{(d-a)(c-a)}{(b-c)(d-b)} $
D) $ \frac{(b-c)(b-d)}{(a-c)(a-d)} $
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Answer:
Correct Answer: B
Solution:
$ a{{\sin }^{2}}x+b{{\cos }^{2}}x=c\Rightarrow (b-a){{\cos }^{2}}x=c-a $
$ \Rightarrow (b-a)=(c-a)(1+{{\tan }^{2}}x) $ $ b{{\sin }^{2}}y+a{{\cos }^{2}}y=d\Rightarrow (a-b){{\cos }^{2}}y=d-b $
$ \Rightarrow (a-b)=(d-b)(1+{{\tan }^{2}}y) $ \ $ {{\tan }^{2}}x=\frac{b-c}{c-a},{{\tan }^{2}}y=\frac{a-d}{d-b} $
$ \therefore \frac{{{\tan }^{2}}x}{{{\tan }^{2}}y}=\frac{(b-c)(d-b)}{(c-a)(a-d)} $ ?..(i) But $ a\tan x=b\tan y, $ i.e., $ \frac{\tan x}{\tan y}=\frac{b}{a} $ ?..(ii) From (i) and (ii), $ \frac{b^{2}}{a^{2}}=\frac{(b-c)(d-b)}{(c-a)(a-d)} $
$ \Rightarrow \frac{a^{2}}{b^{2}}=\frac{(c-a)(a-d)}{(b-c)(d-b)} $ .
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