Trigonometry - Study Mode
[#191] $$frac{{{ ext{tan}} heta + cot heta }}{{{ ext{tan}} heta - cot heta }} = 2,$$ xa0 $$left( {0 leqslant heta leqslant {{90}^ circ }}
ight),$$ xa0 then the value of $$sin heta $$ xa0is?
Correct Answer
(B) $$frac{{sqrt 3 }}{2}$$
Explanation
Solution: $$eqalign{
& frac{{{ ext{tan}} heta + cot heta }}{{{ ext{tan}} heta - cot heta }} = 2 cr
& { ext{By componendo and dividendo}} cr
& Rightarrow frac{{2{ ext{tan}} heta }}{{2{ ext{cos}} heta }} = frac{3}{1} cr
& Rightarrow frac{{sin heta }}{{{ ext{cos}} heta }} imes frac{{sin heta }}{{{ ext{cos}} heta }} = 3 cr
& Rightarrow {sin ^2} heta = 3{ ext{co}}{{ ext{s}}^2} heta cr
& Rightarrow {sin ^2} heta = 3left( {1 - {{sin }^2} heta }
ight) cr
& Rightarrow 4{sin ^2} heta = 3 cr
& Rightarrow {sin ^2} heta Rightarrow frac{3}{4} cr
& Rightarrow { ext{sin }} heta = frac{{sqrt 3 }}{2} cr
& cr
& {x08f{Alternate:}} cr
& Rightarrow frac{{{ ext{tan}} heta + cot heta }}{{{ ext{tan}} heta - cot heta }} = 2 cr
& { ext{By C and D}} cr
& Rightarrow frac{{{ ext{tan}} heta }}{{cot heta }} = frac{3}{1} cr
& Rightarrow { ext{ta}}{{ ext{n}}^2} heta = 3 cr
& Rightarrow { ext{tan}} heta = sqrt 3 cr
& heta = {60^ circ } cr
& Rightarrow sin heta cr
& Rightarrow { ext{sin }}{60^ circ } cr
& Rightarrow frac{{sqrt 3 }}{2} cr} $$
[#192] $$frac{{sin heta + cos heta }}{{{ ext{sin}} heta - cos heta }} = 3,$$ xa0xa0 then the value of $${sin ^4} heta - { ext{co}}{{ ext{s}}^4} heta $$ xa0xa0 is?
Correct Answer
(B) $$frac{3}{5}$$
Explanation
Solution: If in the any question componendo and dividendo already used as $$frac{{a + b}}{{a - b}} = frac{m}{n}$$ If second time you also want to apply componendo dividendo rule then result will be $$eqalign{
& frac{a}{b} = frac{{m + n}}{{m - n}} cr
& Leftrightarrow frac{{sin heta + cos heta }}{{{ ext{sin}} heta - cos heta }} = 3 cr
& Leftrightarrow sin heta + { ext{cos}} heta = 3sin heta - 3{ ext{cos}} heta cr
& Leftrightarrow 2sin heta = 4{ ext{cos}} heta cr
& Leftrightarrow frac{{sin heta }}{{{ ext{cos}} heta }} = frac{2}{1} cr
& Leftrightarrow { ext{tan}} heta = frac{2}{1} = frac{{ ext{P}}}{{ ext{B}}} cr
& left[ { herefore { ext{sin}} heta = frac{{ ext{P}}}{{ ext{H}}} = frac{2}{{sqrt 5 }}{ ext{ and cos}} heta = frac{{ ext{B}}}{{ ext{H}}} = frac{1}{{sqrt 5 }}}
ight] cr} $$ $$eqalign{
& Leftrightarrow {sin ^4} heta - { ext{co}}{{ ext{s}}^4} heta cr
& Leftrightarrow left( {{{sin }^2} heta + { ext{co}}{{ ext{s}}^2} heta }
ight)left( {{{sin }^2} heta - { ext{co}}{{ ext{s}}^2} heta }
ight) cr
& Leftrightarrow 1left( {{{sin }^2} heta - { ext{co}}{{ ext{s}}^2} heta }
ight) cr
& Leftrightarrow {left( {frac{2}{{sqrt 5 }}}
ight)^2} - {left( {frac{1}{{sqrt 5 }}}
ight)^2} cr
& Leftrightarrow frac{4}{5} - frac{1}{5} cr
& Leftrightarrow frac{3}{5} cr} $$
[#193] If tan15° = 2 - $$sqrt 3 ,$$xa0 then the value of tan15° cot75° + tan75° cot15° is?
Correct Answer
(A) 14
Explanation
Solution: $${ ext{ tan 1}}{5^ circ }{ ext{cot 7}}{5^ circ } + { ext{tan 7}}{5^ circ }{ ext{cot 1}}{5^ circ }$$ $$ = { ext{ tan 1}}{5^ circ }{ ext{cot }}left( {{{90}^ circ } - {{15}^ circ }}
ight) + $$ xa0 xa0xa0 $${ ext{tan}}{left( {{{90}^ circ } - 15}
ight)^ circ }$$ xa0xa0 $${ ext{cot1}}{5^ circ }$$ $$eqalign{
& = { ext{ ta}}{{ ext{n}}^2}{ ext{1}}{5^ circ } + { ext{co}}{{ ext{t}}^2}{ ext{1}}{5^ circ } cr
& = { ext{ta}}{{ ext{n}}^2}{15^ circ } + { ext{co}}{{ ext{t}}^2}{15^ circ },.....(i) cr
& left[ {{x08f{Formula}}}
ight] cr
& cot left( {{{90}^ circ } - heta }
ight) = an heta cr
& an left( {{{90}^ circ } - heta }
ight) = cot heta cr
& { ext{Put value of tan1}}{5^ circ } cr
& cot {15^ circ } = frac{1}{{{ ext{tan1}}{5^ circ }}} cr
& cot {15^ circ } = frac{1}{{left( {2 - sqrt 3 }
ight)}} cr
& cot {15^ circ } = frac{1}{{left( {2 - sqrt 3 }
ight)}} imes frac{{left( {2 + sqrt 3 }
ight)}}{{left( {2 + sqrt 3 }
ight)}} cr
& cot {15^ circ } = 2 + sqrt 3 cr
& { ext{Now put value in equation (i)}} cr
& { ext{ tan 1}}{5^ circ } + { ext{cot 1}}{5^ circ } cr
& = {left( {2 - sqrt 3 }
ight)^2} + {left( {2 + sqrt 3 }
ight)^2} cr
& = 4 + 3 - 4sqrt 3 + 4 + 3 + 4sqrt 3 cr
& = 14 cr} $$
[#194] If A = tan11°. tan29°, B = 2cot61°. cot79° then -
Correct Answer
(C) 2A = B
Explanation
Solution: $$eqalign{
& Leftrightarrow frac{{ ext{A}}}{{ ext{B}}} = frac{{{ ext{tan1}}{{ ext{1}}^ circ }{ ext{.tan2}}{{ ext{9}}^ circ }}}{{{ ext{2cot}}{{61}^ circ }.cot {{79}^ circ }}} cr
& Leftrightarrow frac{{ ext{A}}}{{ ext{B}}} = frac{{{ ext{tan1}}{{ ext{1}}^ circ }{ ext{.tan2}}{{ ext{9}}^ circ }}}{{{ ext{2}}left[ {{ ext{cot}}left( {{{90}^ circ } - {{29}^ circ }}
ight).cot left( {{{90}^ circ } - {{11}^ circ }}
ight)}
ight]}} cr
& Leftrightarrow frac{{ ext{A}}}{{ ext{B}}} = frac{{{ ext{tan1}}{{ ext{1}}^ circ }{ ext{.tan2}}{{ ext{9}}^ circ }}}{{{ ext{2tan1}}{{ ext{1}}^ circ }.tan{{29}^ circ }}} cr
& Leftrightarrow frac{{ ext{A}}}{{ ext{B}}} = frac{1}{2} cr
& Leftrightarrow 2{ ext{A}} = { ext{B}} cr} $$
[#195] If $${ ext{tan}}left( {frac{pi }{2} - frac{ heta }{2}}
ight) = sqrt 3 $$ xa0 xa0 the value of cosθ is?
Correct Answer
(C) $$frac{1}{2}$$
Explanation
Solution: $$eqalign{
& { ext{ tan}}left( {frac{pi }{2} - frac{ heta }{2}}
ight) = sqrt 3 cr
& Rightarrow { ext{tan}}left( {{{90}^ circ } - frac{ heta }{2}}
ight) = sqrt 3 ,,left[ {pi = {{180}^ circ }}
ight] cr
& Rightarrow cot frac{ heta }{2} = sqrt 3 cr
& Rightarrow cot frac{ heta }{2} = cot {30^ circ } cr
& Rightarrow frac{ heta }{2} = {30^ circ } cr
& Rightarrow heta = {60^ circ } cr
& Rightarrow cos {60^ circ } = frac{1}{2} cr} $$