11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer If \[\tan x=\frac{2b}{a-c}(a\ne c),\]\[y=a\,{{\cos }^{2}}x+2b\,\sin x\cos x+c\,{{\sin }^{2}}x\]and \[z=a{{\sin }^{2}}x-2b\sin x\cos x+c{{\cos }^{2}}x,\] then

    A) \[y=z\]

    B) \[y+z=a+c\]

    C) \[y-z=a+c\]

    D) \[y-z={{(a-c)}^{2}}+4{{b}^{2}}\]

    Correct Answer: B

    Solution :

    We have, \[y+z=a({{\cos }^{2}}x+{{\sin }^{2}}x)+c({{\sin }^{2}}x+{{\cos }^{2}}x)=a+c\] \[(\therefore \text{solution is (b)}\}\] \[y-z=a({{\cos }^{2}}x-{{\sin }^{2}}x)+4b\sin x\cos x\]\[-c({{\cos }^{2}}x-{{\sin }^{2}}x)\] \[=(a-c)\cos 2x+2b\sin 2x\] \[=(a-c).\,\left( \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)+2b.\left( \frac{2\tan x}{1+{{\tan }^{2}}x} \right)\] \[=(a-c).\left\{ \frac{1-4{{b}^{2}}/{{(a-c)}^{2}}}{1+4{{b}^{2}}/{{(a-c)}^{2}}} \right\}+2b.\left\{ \frac{2.2b/(a-c)}{1+4{{b}^{2}}/{{(a-c)}^{2}}} \right\}\] Since \[\tan x=\frac{2b}{(a-c)}\], \[\therefore y-z=\frac{(a-c).\{{{(a-c)}^{2}}-4{{b}^{2}}\}+8{{b}^{2}}(a-c)}{{{(a-c)}^{2}}+4{{b}^{2}}}\] \[=\frac{(a-c){{(a-c)}^{2}}+4{{b}^{2}}}{\{{{(a-c)}^{2}}+4{{b}^{2}}\}}=(a-c)\] \[\Rightarrow y\ne z\,,\,(\because a\ne c)\].


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