Question

tana+cota=2 where 0° < A < 90° find the value of sin^15A + cos^45 A .

Answer 1

Solution:

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=> tan A + cot A = 2,

=>$Tan A + \frac{1}{Tan A} = 2$

=> $\frac{Tan^2A+1}{Tan A} = 2$

=> $tan^2 A + 1 = 2tan A$

=>[tex]tan^2A -2tanA +1 =0 [/tex]

=> $(tanA - 1)^2 = 0$

=> [tex]Tan A -1 = 0 [/tex]

=> $Tan A = 1$

=> Tan 45° =1

=> Tan A = Tan 45°

=> ∴ A = 45°
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$sin^{15} A + cos ^{45}A$

=> $( \frac{1}{ \sqrt{2} } )^{15} + ( \frac{1}{ \sqrt{2} } )^{45}$

=> $\frac{1}{( \sqrt{2} )^{15}} + \frac{1}{ (\sqrt{2})^{45} }$

=> $\frac{ (\sqrt{2})^{30}+1 }{(( \sqrt{2})^{45} )}$

=> $\frac{2^{15}+1}{2^{22.5}}$

=> $\frac{32768 +1}{4194304 \sqrt{2}}$

=> $\frac{32769}{4194304 \sqrt{2}}$

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                                        Hope it Helps !!