Trigonometry> sir could u pls solve the circled questio...

sir could u pls solve the circled questions in this image
the answers are 3,3,2 (correct options) but pls tell me the procedure

Raghav Rao , 9 Years ago

Grade 12

3 Answers

Ajay

Answer for question 87 . will try to post answers for other question im separate posts…...............................

$Given\quad cot\theta \quad =\quad \frac { 2tan7.5 }{ 1-{ tan }^{ 2 }7.5 } \quad find\quad sin3\theta \\ Using\quad formula\quad for\quad tan2\theta \\ tan15\quad =\quad \frac { 2tan7.5 }{ 1-{ tan }^{ 2 }7.5 } \\ cot\theta \quad =\quad tan15\quad =\quad cot(90-15)\quad =\quad cot75\\ hence\quad \theta \quad =\quad 75\\ Now\quad sin3\theta \quad \quad \quad =\quad sin225\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad sin(180+45)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -sin45\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =-\frac { 1 }{ \sqrt { 2 } }$

Last Activity: 9 Years ago

Ajay

Answer for question 88........................................................................................

$Given\quad x+\frac { 1 }{ x } =2cos20\quad find\quad { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } \\ We\quad know\quad { \left( x+\frac { 1 }{ x } \right) }^{ 3 }\quad \quad =\quad { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3{ x }^{ 2 }\left( \frac { 1 }{ x } \right) +3x\left( \frac { 1 }{ { x }^{ 2 } } \right) \\ or\quad { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } \quad =\quad { \left( x+\frac { 1 }{ x } \right) }^{ 3 }-3\left( x+\frac { 1 }{ x } \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad { (2cos20) }^{ 3 }-3.2cos20\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 2\left( 4{ cos }^{ 3 }20-3cos20 \right) \\ using\quad formula\quad for\quad cos3\theta \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 2cos60\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =1$

Last Activity: 9 Years ago

Ajay

$finaly\quad answer\quad for\quad last\quad question\\ Given\quad sec\theta -cos\theta =1\quad find\quad { tan }^{ 2 }\theta /2\\ \frac { 1 }{ cos\theta } -cos\theta \quad =1\\ { cos }^{ 2 }\theta +cos\theta =1\\ cos\theta \quad =\quad \frac { -1\pm \sqrt { 5 } }{ 2 } \\ Since\quad -1<cos\theta <1\quad and\quad \frac { -1-\sqrt { 5 } }{ 2 } \quad <\quad -1\quad ignoring\quad this\quad root\\ Hence\quad cos\theta \quad =\quad \frac { -1+\sqrt { 5 } }{ 2 } \\ 2{ cos }^{ 2 }\theta /2\quad -\quad 1\quad =\quad \frac { -1+\sqrt { 5 } }{ 2 } \\ { cos }^{ 2 }\theta /2\quad =\quad \frac { \sqrt { 5 } +1 }{ 4 } \\ { sec }^{ 2 }\theta /2\quad =\quad \frac { 4 }{ \sqrt { 5 } +1 } \\ now\quad since\quad { tan }^{ 2 }\theta /2\quad =\quad { sec }^{ 2 }\theta /2-1\\ \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 4 }{ \sqrt { 5 } +1 } -1$