11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer 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

    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) \[\frac{(b-c)\,\,(b-d)}{(a-c)\,\,(a-d)}\]

    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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