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Raj keshri

· started a discussion

· 1 Months ago

doubt (Question: 111615.0 )

will it be true for p=3 and n=2???????????????????????????????

Question:
If θ be the acute angle such that tan θ = 1, then \(\cfrac{1-cos^4\theta}{cos^4\theta}+\cfrac{1+sin^4\theta}{sin^4\theta}\) is equal to:
Options:
A) (24 - 1) + \(\left ( \cfrac {9+2^4}{9} \right )\)
B) 8
C) 4
D) \(\cfrac{2^4-9}{9}+(2^4 + 1)\)
Solution:
Ans: (b)

Since

\(tan\ \theta=1=tan\ \cfrac{\pi}{4}\) 

\(\therefore \theta=\cfrac{\pi}{4}\) 

Hence \(\cfrac{1-cos^cos4\theta}{cos^4\theta}+\cfrac{1+sin^4\theta}{sin^4\theta}\)

\(=\cfrac{1-cos^4\left ( \cfrac {\pi}{4} \right )}{cos^4 \left ( \cfrac {\pi}{4} \right )}\ \ +\ \ \cfrac{1+ sin^4\left ( \cfrac {\pi}{4} \right )}{sin^4 \left ( \cfrac {\pi}{4} \right )}\)


\(=\cfrac{1-\left ( \cfrac {1}{\sqrt{2}} \right )^4}{\left ( \cfrac {1}{\sqrt{2}} \right )^4}\ \ +\ \ =\cfrac{1+\left ( \cfrac {1}{\sqrt{2}} \right )^4}{\left ( \cfrac {1}{\sqrt{2}} \right )^4} \ \ =\ \ \cfrac{1-\cfrac{1}{4}}{\cfrac{1}{4}}4\ \ + \ \ \cfrac{1+\cfrac{1}{4}}{\cfrac{1}{4}}\)

=8

Knowledge Expert

· commented

· 1 Months ago

Dear Student,
We have made some corrections in the solution.
You can check it now.

Team TR

Vijay Sharma

· commented

· 1 Months ago

SAME HERE...this is a weird question because >= includes 2 also which wont satisfy it
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