SATHEE: Quantitative Aptitude Ques 1777

Quantitative Aptitude Ques 1777

Question: If $ \cos x+{{\cos }^{2}}x=1, $ then the numerical value of $ ({{\sin }^{12}}x+3{{\sin }^{10}}x+3{{\sin }^{8}}x+{{\sin }^{6}}x-1) $ is

Options:

A) $ 0 $

B) $ 1 $

C) $ -1 $

D) $ 2 $

Show Answer

Answer:

Correct Answer: A

Solution:

Given, $ \cos x+{{\cos }^{2}}x=1 $

$ \Rightarrow $ $ cox=1-{{\cos }^{2}}x $

$ \Rightarrow $ $ \cos x={{\sin }^{2}}x $ (i) Now, again $ \cos x+{{\cos }^{2}}x=1 $ On cubing both side, we get $ {{(\cos x+{{\cos }^{2}}x)}^{3}}={{(1)}^{3}} $

$ \Rightarrow $ $ {{\cos }^{3}}x+{{({{\cos }^{2}}x)}^{3}}+3{{\cos }^{2}}x{{\cos }^{2}}x $ $ +3\cos x{{\cos }^{4}}x=1 $

$ \Rightarrow $ $ {{\cos }^{3}}x+{{\cos }^{6}}x+3{{\cos }^{4}}x+3{{\cos }^{5}}x=1 $

$ \Rightarrow $ $ {{\sin }^{6}}x+{{\sin }^{12}}x+3{{\sin }^{8}}x+3{{\sin }^{10}}x=1 $ [from Eq. (i)]

$ \therefore $ $ {{\sin }^{12}}x+3{{\sin }^{10}}x+3{{\sin }^{8}}x+{{\sin }^{6}}x-1=0 $

Quantitative Aptitude Ques 1776

Quantitative Aptitude Ques 1778

Quantitative Aptitude Ques 1777

Question: If $ \cos x+{{\cos }^{2}}x=1, $ then the numerical value of $ ({{\sin }^{12}}x+3{{\sin }^{10}}x+3{{\sin }^{8}}x+{{\sin }^{6}}x-1) $ is

Options:

Answer:

Solution:

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