Trigonometry - Study Mode
[#166] If tan 2 θ = 1 - e 2 , then the value of secθ + tan 3 θ.cosecθ is?
Correct Answer
(D) $${left( {2 - {e^2}}
ight)^{frac{3}{2}}}$$
Explanation
Solution: $$eqalign{
& { ext{ta}}{{ ext{n}}^2} heta = 1 - {e^2} cr
& herefore sec heta + { ext{ta}}{{ ext{n}}^3} heta . { ext{cosec}} heta cr
& Rightarrow sec heta + { ext{ta}}{{ ext{n}}^2} heta .{ ext{tan}} heta . { ext{cosec}} heta cr
& Rightarrow sec heta + { ext{ta}}{{ ext{n}}^2} heta .frac{{sin heta }}{{{ ext{cos}} heta }}.frac{1}{{sin heta }} cr
& Rightarrow sec heta + { ext{ta}}{{ ext{n}}^2} heta .sec heta cr
& Rightarrow sec heta left( {1 + { ext{ta}}{{ ext{n}}^2} heta }
ight) cr
& Rightarrow sqrt {1 + { ext{ta}}{{ ext{n}}^2} heta } .left( {1 + { ext{ta}}{{ ext{n}}^2} heta }
ight) cr
& Rightarrow {left( {1 + { ext{ta}}{{ ext{n}}^2} heta }
ight)^{frac{3}{2}}} cr
& Rightarrow {left( {1 + 1 - {e^2}}
ight)^{frac{3}{2}}} cr
& Rightarrow {left( {2 - {e^2}}
ight)^{frac{3}{2}}} cr} $$
[#167] If $$frac{{{ ext{cos }}alpha }}{{{ ext{cos }}x08eta }} = a$$ xa0 and $$frac{{{ ext{sin }}alpha }}{{{ ext{sin }}x08eta }} = b{ ext{,}}$$ xa0 then the value of $${sin ^2}x08eta $$ xa0in terms of a and b is?
Correct Answer
(C) $$frac{{{a^2} - 1}}{{{a^2} - {b^2}}}$$
Explanation
Solution: $$eqalign{
& frac{{{ ext{cos }}alpha }}{{{ ext{cos }}x08eta }} = a{ ext{ }} cr
& Rightarrow cos { ext{ }}alpha = a{ ext{ }}cos { ext{ }}x08eta cr
& { ext{On squaring both sides}} cr
& {cos ^2}alpha = {a^2}{cos ^2}x08eta cr
& Rightarrow 1 - {sin ^2}alpha = {a^2}left( {1 - {{sin }^2}x08eta }
ight)....(i) cr
& { ext{Again, }}sin alpha = { ext{ }}bsin x08eta cr
& { ext{Squaring both sides}} cr
& Rightarrow {sin ^2}alpha = { ext{ }}{b^2}{sin ^2}x08eta cr
& { ext{Put the value of }}{sin ^2}alpha { ext{ in equation (i)}} cr
& Rightarrow { ext{1}} - {b^2}{sin ^2}x08eta = {a^2} - {a^2}si{n^2}x08eta cr
& Rightarrow {a^2} - 1 = {a^2}si{n^2}x08eta - {b^2}si{n^2}x08eta cr
& Rightarrow {a^2} - 1 = si{n^2}x08eta left( {{a^2} - {b^2}}
ight) cr
& Rightarrow si{n^2}x08eta = frac{{{a^2} - 1}}{{{a^2} - {b^2}}} cr} $$
[#168] sinθ = 0.7 then cosθ, 0 ≤ θ < 90° is?
Correct Answer
(C) $$sqrt {0.51} $$
Explanation
Solution: $$eqalign{
& sin heta = 0.7 cr
& Rightarrow {sin ^2} heta + { ext{co}}{{ ext{s}}^2} heta = 1 cr
& Rightarrow {left( {0.7}
ight)^2} + { ext{co}}{{ ext{s}}^2} heta = 1 cr
& Rightarrow 0.49 + { ext{co}}{{ ext{s}}^2} heta = 1 cr
& Rightarrow { ext{co}}{{ ext{s}}^2} heta = 1 - 0.49 cr
& Rightarrow { ext{cos}} heta = sqrt {0.51} cr} $$
[#169] If $${ ext{2}}sin heta + { ext{cos}} heta = frac{7}{3}{ ext{,}}$$ xa0xa0 then the value of $$left( {{ ext{ta}}{{ ext{n}}^2} heta - {{sec }^2} heta }
ight)$$ xa0 is?
Correct Answer
(B) -1
Explanation
Solution: $$eqalign{
& { ext{2}}sin heta + { ext{cos}} heta = frac{7}{3} cr
& Rightarrow left( {{ ext{ta}}{{ ext{n}}^2} heta - { ext{se}}{{ ext{c}}^2} heta }
ight) cr
& Rightarrow left( {{{sec }^2} heta - 1 - { ext{se}}{{ ext{c}}^2} heta }
ight) cr
& ,,,,,,,,,,,left[ {x08ecause 1 + { ext{ta}}{{ ext{n}}^2} heta = {{sec }^2} heta }
ight] cr
& Rightarrow - 1 cr} $$
[#170] The value of $${sec ^2} heta $$ xa0- $$frac{{{{sin }^2} heta - 2{{sin }^4} heta }}{{{ ext{2co}}{{ ext{s}}^4} heta - { ext{co}}{{ ext{s}}^2} heta }}$$ xa0 is?
Correct Answer
(A) 1
Explanation
Solution: $$eqalign{
& {sec ^2} heta - frac{{{{sin }^2} heta - 2{{sin }^4} heta }}{{{ ext{2co}}{{ ext{s}}^4} heta - { ext{co}}{{ ext{s}}^2} heta }} cr
& Rightarrow {sec ^2} heta - frac{{{{sin }^2} heta left( {1 - 2{{sin }^2} heta }
ight)}}{{{ ext{co}}{{ ext{s}}^2} heta left( {{ ext{2co}}{{ ext{s}}^2} heta - 1}
ight)}} cr
& left[ {{ ext{co}}{{ ext{s}}^2} heta - {{sin }^2} heta = 2{ ext{co}}{{ ext{s}}^2} heta - 1 = 1 - 2{{sin }^2} heta }
ight] cr
& Rightarrow {sec ^2} heta - { ext{ta}}{{ ext{n}}^2} heta cr
& Rightarrow 1 cr} $$