Trigonometry - Study Mode

[#216] Find the value of sin 2 25° + sin 2 65° + coses 2 57° - tan 2 33° = ?
Correct Answer

(B) 2

Explanation

Solution: $${sin ^2}{25^ circ } + {sin ^2}{65^ circ } + { ext{cose}}{{ ext{c}}^2}{57^ circ } - { ext{ta}}{{ ext{n}}^2}{33^ circ }$$ $$ Rightarrow left( {{{sin }^2}{{25}^ circ } + {{cos }^2}{{25}^ circ }}
ight) + $$ xa0 xa0 $$left( {{ ext{se}}{{ ext{c}}^2}{{33}^ circ } - { ext{ta}}{{ ext{n}}^2}{{33}^ circ }}
ight)$$ $$eqalign{
& Rightarrow 1 + 1 cr
& Rightarrow 2 cr
& {x08f{Note:}} cr
& {sin ^2}{65^ circ } = {sin ^2}left( {{{90}^ circ } - {{25}^ circ }}
ight) cr
& ,,,,,,,,,,,,,,,,,,,, = { ext{co}}{{ ext{s}}^2}{25^ circ } cr
& { ext{cose}}{{ ext{c}}^2}{57^ circ } = {operatorname{cosec} ^2}left( {{{90}^ circ } - {{33}^ circ }}
ight) cr
& ,,,,,,,,,,,,,,,,,,,,,,,,, = { ext{se}}{{ ext{c}}^2}{33^ circ } cr} $$

[#217] If $$6{sin ^4} heta + 3{cos ^4} heta = 2{ ext{,}}$$ xa0 xa0 then the value of $${left[ {7{{operatorname{cosec} }^6} heta + 8{{sec }^6} heta }
ight]^{frac{1}{3}}}$$ xa0 xa0 is?
Correct Answer

(D) 6

Explanation

Solution: 6sin 4 θ + 3cos 4 θ = 2 ⇒ 6sin 4 θ + 3(1 - sin 2 θ ) 2 = 2 ⇒ 6sin 4 θ + 3 + 3sin 4 θ - 6sin 2 θ = 2 ⇒ 9sin 4 θ - 6sin 2 θ + 1 = 0 ⇒ (3sin 2 θ -1) 2 = 0 ⇒ 3sin 2 θ = 1 ⇒ sinθ = $$frac{1}{{sqrt 3 }} $$ Now, $${left[ {7{{operatorname{cosec} }^6} heta + 8{{sec }^6} heta }
ight]^{frac{1}{3}}}$$ $$eqalign{
& = {left[ {7 imes {{(sqrt 3 )}^6} + 8{{left( {frac{{sqrt 3 }}{{sqrt 2 }}}
ight)}^6}}
ight]^{frac{1}{3}}} cr
& = {left[ {7 imes 27 + 8 imes frac{{27}}{8}}
ight]^{frac{1}{3}}} cr
& = {left( {8 imes 27}
ight)^{frac{1}{3}}} cr
& = 6 cr} $$

[#218] $$frac{{2sin heta }}{{cos heta left( {1 + { ext{ta}}{{ ext{n}}^2} heta }
ight)}}$$ xa0 xa0 simplifies to?
Correct Answer

(C) sin2θ

Explanation

Solution: $$eqalign{
& = frac{{2sin heta }}{{cos heta left( {1 + { ext{ta}}{{ ext{n}}^2} heta }
ight)}} cr
& = frac{{2{ ext{tan}} heta }}{{1 + { ext{ta}}{{ ext{n}}^2} heta }} cr
& = sin2 heta cr} $$

[#219] If $$ an heta = frac{3}{4}{ ext{,}}$$ xa0 find the value of $${ ext{cos2}} heta $$xa0 ?
Correct Answer

(C) $$frac{7}{{25}}$$

Explanation

Solution: $$eqalign{
& an heta = frac{3}{4} cr
& { ext{cos2}} heta = frac{{1 - { ext{ta}}{{ ext{n}}^2} heta }}{{1 + { ext{ta}}{{ ext{n}}^2} heta }} cr
& ,,,,,,,,,,,,,,,,, = frac{{1 - frac{9}{{16}}}}{{1 + frac{9}{{16}}}} cr
& ,,,,,,,,,,,,,,,,,, = frac{{frac{7}{{16}}}}{{frac{{25}}{{16}}}} cr
& { ext{cos2}} heta = frac{7}{{25}} cr} $$

[#220] The value of $$frac{{{ ext{sin 6}}{5^ circ }}}{{cos {{25}^ circ }}}$$ xa0is?
Correct Answer

(B) 1

Explanation

Solution: $$eqalign{
& = frac{{{ ext{sin 6}}{5^ circ }}}{{cos {{25}^ circ }}} cr
& = frac{{{ ext{sin }}left( {{{90}^ circ } - { ext{6}}{5^ circ }}
ight)}}{{cos {{25}^ circ }}} cr
& left[ {sin left( {{{90}^ circ } - heta }
ight) = cos heta }
ight] cr
& = frac{{cos {{25}^ circ }}}{{cos {{25}^ circ }}} cr
& = 1 cr} $$