Math, asked by BrainlyHelper, 1 year ago

निम्नलिखित के मान निकालिए:
(i) sin60^{o} cos30^{o} + sin30^{o}cos 60^{o}
(ii)  2tan ^{2} 45^{o} + cos ^{2} 30^{o}- sin ^{2} 60^{o}
(iii)  \frac{cos 45^{o}}{sec 30^{o}+cosec 30^{o}} ,
(iv)  \frac{sin 30^{o}+tan 45^{o}-cosec 60^{o}}{sec 30^{o}+cos 60^{o}+cot 45^{o}} ,
(v)  \frac{ 5 cos^{2} 60^{o}+4 sec^{2} 30^{o}-tan^{2} 45^{o}}{sin^{2}30^{o}+cos^{2} 30^{o}}

Answers

Answered by hukam0685
0
निम्नलिखित के मान निकालिए:

(i) sin60^{o} cos30^{o} + sin30^{o}cos 60^{o}

 \sin(60)  \cos(30)  +  \sin(30)  \cos(60)  \\  \\  =  \frac{ \sqrt{3} }{2}  \times  \frac{ \sqrt{3} }{2}  +  \frac{1}{2}  \times  \frac{1}{2}  \\  \\  =  \frac{3}{4}  +  \frac{1}{4}  \\  \\  =    \frac{4}{4}  \\  \\  = 1


(ii)  2tan ^{2} 45^{o} + cos ^{2} 30^{o}- sin ^{2} 60^{o}

2 \times ( {1)}^{2}  + ( { \frac{ \sqrt{3} }{2} })^{2}  -  ({ \frac{ \sqrt{3} }{2} })^{2}  \\  \\  = 2
(iii)  \frac{cos 45^{o}}{sec 30^{o}+cosec 30^{o}} ,

 \frac{ \frac{1}{ \sqrt{2} } }{ \frac{2}{ \sqrt{3} }   + 2}  \\  \\  =  \frac{ \frac{1}{ \sqrt{2} } }{ \frac{2  + 2 \sqrt{3} }{ \sqrt{3} }  }  \\ \\  =  \frac{ \sqrt{3} }{ \sqrt{2}(2 + 2 \sqrt{3} ) }  \\  \\  =  \frac{ \sqrt{3} }{2 \sqrt{2} (1 +  \sqrt{3}) }  \times  \frac{(1 -  \sqrt{3} )}{(1 -  \sqrt{3}) }  \\  \\  =  \frac{ \sqrt{3}(1 -  \sqrt{3}  )}{2 \sqrt{2} (1 - 3)}  \\  \\  =  \frac{3 -  \sqrt{3} }{4 \sqrt{2} }  \times  \frac{ \sqrt{2} }{ \sqrt{2} }  \\  \\  =  \frac{ \sqrt{2}(3 -  \sqrt{3})  }{8}

(iv)  \frac{sin 30^{o}+tan 45^{o}-cosec 60^{o}}{sec 30^{o}+cos 60^{o}+cot 45^{o}} ,

 \frac{ \frac{1}{2} + 1 -  \frac{2}{ \sqrt{3} }  }{ \frac{2}{ \sqrt{3} }  +  \frac{1}{2}  + 1}  \\  \\  =  \frac{ \sqrt{3} + 2 \sqrt{3}  - 4 }{4 +  \sqrt{3}  +2 \sqrt{3}  }  \\  \\  =  \frac{3 \sqrt{3} - 4 }{3 \sqrt{3}  + 4}  \\  \\  = \frac{3 \sqrt{3} - 4 }{3 \sqrt{3}  + 4}  \times \frac{3 \sqrt{3} - 4 }{3 \sqrt{3}   -  4}  \\  \\  \\   =  \frac{ {(3 \sqrt{3}  - 4)}^{2} }{27 - 16}  \\  \\  =  \frac{27  + 16 - 24 \sqrt{3} }{11}  \\  \\  =  \frac{43 - 24 \sqrt{3} }{11}



(v)  \frac{ 5 cos^{2} 60^{o}+4 sec^{2} 30^{o}-tan^{2} 45^{o}}{sin^{2}30^{o}+cos^{2} 30^{o}}

 \frac{5 \times ( { \frac{1}{2} })^{2} + 4( { \frac{ 2 }{ \sqrt{3} } })^{2}  - 1 }{ ( { \frac{1}{2} }  )^{2} +  ({ \frac{ \sqrt{3} }{2} })^{2}  }  \\  \\  =  \frac{ \frac{5}{4} +  \frac{16}{3}  - 1 }{ \frac{1}{4}  +  \frac{3}{4} }  \\  \\  =  \frac{ \frac{15 + 64 - 12}{12} }{1}  \\  \\  =  \frac{67}{12}

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