Trigonometry - Study Mode
[#226] If $$x = frac{{2sin heta }}{{1 + cos heta + sin heta }},$$ xa0 xa0 then the value of $$frac{{1 - cos heta + sin heta }}{{1 + sin heta }}$$ xa0 is
Correct Answer
(B) $$x$$
Explanation
Solution: $$eqalign{
& x = frac{{2sin heta }}{{1 + cos heta + sin heta }} cr
& { ext{Let's put the value of }} heta = {45^ circ } cr
& { ext{So, }}x = frac{{2 imes frac{1}{{sqrt 2 }}}}{{1 + frac{1}{{sqrt 2 }} + frac{1}{{sqrt 2 }}}} cr
& left{ {{ ext{Because }}sin {{45}^ circ } = frac{1}{{sqrt 2 }}{ ext{ and}}cos {{45}^ circ } = frac{1}{{sqrt 2 }}}
ight} cr
& x = frac{{sqrt 2 }}{{frac{{left( {sqrt 2 + 2}
ight)}}{{sqrt 2 }}}} cr
& x = frac{{sqrt 2 }}{{left( {1 + sqrt 2 }
ight)}} cr
& { ext{Now, check in the required value}} cr
& frac{{1 - cos heta + sin heta }}{{1 + sin heta }} cr
& = frac{{1 - frac{1}{{sqrt 2 }} + frac{1}{{sqrt 2 }}}}{{1 + frac{1}{{sqrt 2 }}}} cr
& = frac{{sqrt 2 }}{{left( {sqrt 2 + 1}
ight)}} cr
& { ext{This values is same as }}x. cr
& { ext{Hence answer is }}x. cr} $$
[#227] If tan4θ = cot(40° - 2θ), then θ is equal to:
Correct Answer
(B) 25°
Explanation
Solution: tan4θ = cot(40° - 2θ) tan4θ = cot(90° - 40° + 2θ) tan4θ = tan(50° + 2θ) 4θ = 50° + 2θ 2θ = 50° θ = 25°
[#228] What is the value of $$frac{{32{{cos }^6}x - 48{{cos }^4}x + 18{{cos }^2}x - 1}}{{4sin x,cos x,sin left( {60 - x}
ight)cos left( {60 - x}
ight)sin left( {60 + x}
ight)cos left( {60 + x}
ight)}}?$$
Correct Answer
(C) 8cot6x
Explanation
Solution: $$eqalign{
& frac{{32{{cos }^6}x - 48{{cos }^4}x + 18{{cos }^2}x - 1}}{{4sin x, imes cos x imes ,sin left( {60 - x}
ight) imes cos left( {60 - x}
ight) imes sin left( {60 + x}
ight) imes cos left( {60 + x}
ight)}} cr
& { ext{Adding }} + 4,{ ext{and }} - 4, cr
& Rightarrow 32{cos ^6}x - 4 - 48{cos ^4}x + 18{cos ^2}x + 3 cr
& Rightarrow 32{cos ^6}x - 4 - 48{cos ^4}x + 24{cos ^2}x - 6{cos ^2}x + 3 cr
& Rightarrow 32{cos ^6}x - 4 - 48{cos ^4}x + 24{cos ^2}x - 3cos 2x cr
& Rightarrow 4{left( {2{{cos }^2}x - 1}
ight)^3} - 3cos 2x cr
& left[ { herefore ,cos 2x = 2{{cos }^2}x - 1}
ight] cr
& Rightarrow 4{cos ^3}2x - 3cos 2x cr
& Rightarrow cos 6x cr} $$ [left[ x08egin{align}
& x08ecause ,cos 3x o 4{{cos }^{3}}x-3cos x \
& ext{and} \
& x08ecause ,cos 6x o 4{{cos }^{3}}2x-3cos 2x \
end{align}
ight]] $$eqalign{
& { ext{Now, }}4sin x,.cos x.,sin left( {60 - x}
ight).cos left( {60 - x}
ight).sin left( {60 + x}
ight).cos left( {60 + x}
ight) cr
& left[ { herefore ,sin x.sin left( {60 - x}
ight).sin left( {60 + x}
ight) = frac{1}{4}sin 3x}
ight] cr
& left[ { herefore ,cos x.cos left( {60 - x}
ight).cos left( {60 + x}
ight) = frac{1}{4}cos 3x}
ight] cr
& = frac{{cos 6x}}{{frac{1}{4}sin 3x.cos 3x}} cr
& = frac{2}{2} imes frac{{4cos 6x}}{{sin 3x.cos 3x}} cr
& = frac{{8cos 6x}}{{2sin 3x.cos 3x}} cr
& = frac{{8cos 6x}}{{sin 6x}} cr
& = 8cot 6x cr} $$
[#229] $$left( {frac{{{{ an }^3} heta }}{{{{sec }^2} heta }} + frac{{{{cot }^3} heta }}{{{ ext{cose}}{{ ext{c}}^2} heta }} + 2sin heta cos heta }
ight)$$ xa0 xa0 xa0÷ (1 + cosec 2 θ + tan 2 θ), 0° < θ < 90°, is equal to:
Correct Answer
(C) sinθcosθ
Explanation
Solution: $$eqalign{
& left( {frac{{{{ an }^3} heta }}{{{{sec }^2} heta }} + frac{{{{cot }^3} heta }}{{{ ext{cose}}{{ ext{c}}^2}, heta }} + 2sin heta cos heta }
ight) div left( {1 + { ext{cose}}{{ ext{c}}^2} heta + {{ an }^2} heta }
ight) cr
& = left( {frac{{{{sin }^3} heta }}{{cos heta }} + frac{{{{cos }^3} heta }}{{{ ext{sin}}, heta }} + 2sin heta cos heta }
ight) div left( {{ ext{cose}}{{ ext{c}}^2} heta + {{sec }^2} heta }
ight) cr
& = left( {frac{{{{sin }^4} heta + {{cos }^4} heta + 2{{sin }^2} heta {{cos }^2} heta }}{{sin heta cos heta }}}
ight) div left( {frac{1}{{{{sin }^2} heta }} + frac{1}{{{{cos }^2} heta }}}
ight) cr
& = left( {frac{{{{left( {{{sin }^2} heta + {{cos }^2} heta }
ight)}^2}}}{{sin heta cos heta }}}
ight) div left( {frac{{{{sin }^2} heta + {{cos }^2} heta }}{{{{sin }^2} heta {{cos }^2} heta }}}
ight) cr
& = left( {frac{1}{{sin heta cos heta }}}
ight) imes left( {frac{{{{sin }^2} heta {{cos }^2} heta }}{1}}
ight) cr
& = sin heta cos heta cr} $$
[#230] The expression $$frac{{{{left( {1 - sin heta + cos heta }
ight)}^2}left( {1 - cos heta }
ight){{sec }^3} heta ,{ ext{cose}}{{ ext{c}}^2} heta }}{{left( {sec heta - an heta }
ight)left( { an heta + cot heta }
ight)}},$$ xa0 xa0 xa0 xa00° < θ < 90°, is equal to:
Correct Answer
(A) 2tanθ
Explanation
Solution: $$eqalign{
& frac{{{{left( {1 - sin heta + cos heta }
ight)}^2}left( {1 - cos heta }
ight){{sec }^3} heta { ext{cose}}{{ ext{c}}^2} heta }}{{left( {sec heta - an heta }
ight)left( { an heta + cot heta }
ight)}} cr
& { ext{put }} heta = {30^ circ } cr
& = frac{{{{left( {1 - sin {{30}^ circ } + cos {{30}^ circ }}
ight)}^2}left( {1 - cos {{30}^ circ }}
ight){{sec }^3}{{30}^ circ }{ ext{cose}}{{ ext{c}}^2}{{30}^ circ }}}{{left( {sec {{30}^ circ } - an {{30}^ circ }}
ight)left( { an {{30}^ circ } + cot {{30}^ circ }}
ight)}} cr
& = frac{{{{left( {1 - frac{1}{2} + frac{{sqrt 3 }}{2}}
ight)}^2}left( {1 - frac{{sqrt 3 }}{2}}
ight){{left( {frac{2}{{sqrt 3 }}}
ight)}^3}{{left( 2
ight)}^2}}}{{left( {frac{2}{{sqrt 3 }} - frac{1}{{sqrt 3 }}}
ight)left( {frac{1}{{sqrt 3 }} + sqrt 3 }
ight)}} cr
& = frac{{{{left( {frac{{1 + sqrt 3 }}{2}}
ight)}^2}left( {frac{{2 - sqrt 3 }}{2}}
ight){{left( {frac{2}{{sqrt 3 }}}
ight)}^3}{{left( 2
ight)}^2}}}{{left( {frac{1}{{sqrt 3 }}}
ight)left( {frac{{1 + 3}}{{sqrt 3 }}}
ight)}} cr
& = frac{{frac{{4 + 2sqrt 3 }}{4} imes frac{{2 - sqrt 3 }}{2} imes frac{8}{{3sqrt 3 }} imes 4}}{{left( {frac{1}{{sqrt 3 }}}
ight)left( {frac{4}{{sqrt 3 }}}
ight)}} cr
& = frac{{left[ {{2^2} - {{left( {sqrt 3 }
ight)}^2}}
ight] imes frac{8}{{3sqrt 3 }}}}{{frac{4}{3}}} cr
& = frac{{frac{8}{{3sqrt 3 }}}}{{frac{4}{3}}} cr
& = frac{2}{{sqrt 3 }} cr
& { ext{Option A: }}2 an {30^ circ } = frac{2}{{sqrt 3 }} cr
& { ext{Option B: }}cot {30^ circ } = sqrt 3 cr
& { ext{Option C: }}sin {30^ circ } = frac{1}{2} cr
& { ext{Option D: }}2cos {30^ circ } = 2 imes frac{{sqrt 3 }}{2} = sqrt 3 cr
& { ext{Only option A is answer}}{ ext{.}} cr} $$