Trigonometry - Study Mode
[#71] If sinθ = 4cosθ, then what is the value of sinθcosθ?
Correct Answer
(C) $$frac{4}{{17}}$$
Explanation
Solution: $$eqalign{
& sin heta = 4cos heta cr
& an heta = 4 cr
& sin heta .cos heta = frac{4}{{sqrt {17} }} imes frac{1}{{sqrt {17} }} = frac{4}{{17}} cr} $$
[#72] What is the value of $$frac{{2left( {1 - {{sin }^2} heta }
ight){ ext{cose}}{{ ext{c}}^2} heta }}{{{{cot }^2} heta left( {1 + {{ an }^2} heta }
ight)}} - 1$$
Correct Answer
(D) cos2θ
Explanation
Solution: $$eqalign{
& frac{{2left( {1 - {{sin }^2} heta }
ight){ ext{cose}}{{ ext{c}}^2} heta }}{{{{cot }^2} heta left( {1 + {{ an }^2} heta }
ight)}} - 1 cr
& = frac{{2{{cos }^2} heta { ext{cose}}{{ ext{c}}^2} heta }}{{{{cot }^2} heta {{sec }^2} heta }} - 1 cr
& = frac{{2{{cos }^2} heta {{sin }^2} heta }}{{{{sin }^2} heta {{cos }^2} heta {{sec }^2} heta }} - 1 cr
& = 2{cos ^2} heta - 1 cr
& = cos 2 heta cr} $$
[#73] What is the value of $$frac{{{{left[ { an left( {{{90}^ circ } - A}
ight) + cot left( {{{90}^ circ } - A}
ight)}
ight]}^2}}}{{left[ {2{{sec }^2}left( {{{90}^ circ } - 2A}
ight)}
ight]}}?$$
Correct Answer
(C) 2
Explanation
Solution: $$eqalign{
& frac{{{{left[ { an left( {{{90}^ circ } - A}
ight) + cot left( {{{90}^ circ } - A}
ight)}
ight]}^2}}}{{left[ {2{{sec }^2}left( {{{90}^ circ } - 2A}
ight)}
ight]}} cr
& { ext{By putting }}A = {45^ circ } cr
& Rightarrow frac{{{{left[ { an left( {{{90}^ circ } - {{45}^ circ }}
ight) + cot left( {{{90}^ circ } - {{45}^ circ }}
ight)}
ight]}^2}}}{{left[ {2{{sec }^2}left( {{{90}^ circ } - 2 imes {{45}^ circ }}
ight)}
ight]}} cr
& Rightarrow frac{{{{left[ { an {{45}^ circ } + cot {{45}^ circ }}
ight]}^2}}}{{2{{sec }^2}left( {{{90}^ circ } - {{90}^ circ }}
ight)}} cr
& Rightarrow frac{{{{left[ {1 + 1}
ight]}^2}}}{{2{{sec }^2}{0^ circ }}} cr
& Rightarrow frac{4}{{2 imes 1}} cr
& Rightarrow 2 cr} $$
[#74] $${left( {frac{{sin heta - 2{{sin }^3} heta }}{{2{{cos }^3} - cos heta }}}
ight)^2} + 1,$$ xa0 xa0 θ ≠ 45° is equal to:
Correct Answer
(B) sec 2 θ
Explanation
Solution: $$eqalign{
& {left( {frac{{sin heta - 2{{sin }^3} heta }}{{2{{cos }^3} - cos heta }}}
ight)^2} + 1 cr
& = frac{{{{sin }^2} heta }}{{{{cos }^2} heta }}{left( {frac{{{{cos }^2} heta }}{{{{cos }^2} heta }}}
ight)^2} + 1 cr
& = { an ^2} heta + 1 cr
& = {sec ^2} heta cr} $$
[#75] If $$frac{{1 + sin heta }}{{1 - sin heta }} = frac{{{p^2}}}{{{q^2}}},$$ xa0 xa0then secθ is equal to:
Correct Answer
(B) $$frac{1}{2}left( {frac{q}{p} + frac{p}{q}}
ight)$$
Explanation
Solution: $$eqalign{
& frac{{1 + sin heta }}{{1 - sin heta }} = frac{{{p^2}}}{{{q^2}}} cr
& { ext{Apply componendo and dividendo}} cr
& frac{1}{{sin heta }} = frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}} cr
& sin heta = frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}} cr} $$ $$eqalign{
& AB = sqrt {{{left( {{p^2} + {q^2}}
ight)}^2} - {{left( {{p^2} - {q^2}}
ight)}^2}} cr
& AB = sqrt {4{p^2}{q^2}} cr
& AB = 2pq cr
& sec heta = frac{{{p^2} + {q^2}}}{{2pq}} cr
& sec heta = frac{1}{2}left[ {frac{p}{q} + frac{q}{p}}
ight] cr} $$