Math, asked by bhaveshgajengi, 7 months ago

tan A = 1/√3 then 7 sin²A + 3cos² A =?
who anwers fast & correctly I will mark as brainliest + Thanks+5.0 vote + i will follow them​​

Answers

Answered by kunal2030
22

Answer:

tanA = 1/root

tanA=tan30

A=30

=7sin^2A+3cos^2A

=7(sin30)^2+3(cos30)^2

=7(1/2)^2+3(root3/2)^2

=7×1/4+3×3/4

=7/4+9/4

=16/4

=4 answer

Answered by TheSentinel
43

{\purple{\underline{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}}} \\ \\

{\rm{ \tan(A) = \dfrac{1}{ \sqrt{3}}\:then\: 7{ \sin}^{2}(A)  + 3{ \cos }^{2} (A) = ?}}

_________________________________________

{\purple{\underline{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}}} \\ \\

{\red{\rm{7{ \sin}^{2}(A)  + 3{ \cos }^{2} (A) = 4}}}

_________________________________________

{\large{\underline{\pink{\sf{Given:}}}}} \\ \\

{\rm{ \tan(A) = \dfrac{1}{ \sqrt{3}}}}

_________________________________________

{\large{\underline{\blue{To \ Find}}}} \\ \\

{\red{\rm{7{ \sin}^{2} (A) + 3{ \cos }^{2} (A) = ?}}}

_________________________________________

{\purple{\underline{\underline{\red{\boxed{\boxed{\blue{\star{\sf Solution:}}}}}}}}} \\ \\

{\rm{We\:are\:Given,}}

{\rm{ \tan(A) = \dfrac{1}{ \sqrt{3}}}} \\ \\

{\therefore{\rm{ A = { \tan}^{-1}( \dfrac{1}{ \sqrt{3}}) }}} \\ \\

{\longrightarrow{\rm{ But , \  { \tan}^{-1}( \dfrac{1}{ \sqrt{3}}) = 30 \degree }}} \\ \\

{\therefore{\orange{\boxed{\green{ A = 30 \degree}}}}} \\ \\

{\rm{ Now,}} \\

{\rm{7{ \sin }^{2} (A) + 3{ \cos }^{2} (A) }} \\ \\

{\rm{Put\:A= 30 \degree \:in\:above\: equation}} \\ \\

{\therefore{\rm{ We\:get,}}} \\ \\

{\implies{\rm{7 { \sin}^{2} (30 \degree) + 3 { \cos}^{2} (30 \degree)}}} \\ \\

{\rm{but\: {\boxed{\star{ \sin(30 \degree) = \dfrac{1}{2}}}}}} \\ \\

{\rm{and\:{\boxed{\star{ \cos(30 \degree) = \dfrac{ \sqrt{3}}{2}}}}}} \\ \\

{\implies{\rm{7 { [ \dfrac{1}{2} ]}^{2} + 3{ [  \dfrac{ \sqrt{3}}{2}]}^{2}}}} \\ \\

{\implies{\rm{ 7 [ \dfrac{1}{4}]+3 [ \dfrac{ 3}{4}]}}} \\ \\

{\implies{\rm{  \dfrac{7}{4}+ [ \dfrac{ 3 \times 3}{4}]}}} \\ \\

{\implies{\rm{  \dfrac{7}{4}+  \dfrac{ 9}{4}}}} \\ \\

{\implies{\rm{  \dfrac{7 + 9}{4}}}} \\ \\

{\implies{\rm{  \dfrac{16}{4}}}} \\ \\

{\implies{\rm{ 4 }}} \\ \\

{\pink{\therefore{\purple{\boxed{\red{\boxed{\blue{7{ \sin}^{2}(A)  + 3{ \cos }^{2} (A) = 4 }}}}}}}}


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