Trigonometry - Study Mode

[#126] If 4sin 2 θ = 3(1 + cosθ), 0° < θ < 90°, then what is the value of (2tanθ + 4sinθ - secθ)?
Correct Answer

(A) $$3sqrt {15} - 4$$

Explanation

Solution: 4sin 2 θ = 3(1 + cosθ) 4(1 - cos 2 θ) = 3 + 3cosθ 4 - 4cos 2 θ = 3 + 3cosθ 4cos 2 θ + 3cosθ - 1 = 0 4cos 2 θ + 4cosθ - cosθ - 1 = 0 4cosθ(cosθ + 1) - 1(cosθ + 1) = 0 (4cosθ -1)(cosθ + 1) = 0 4cosθ - 1 = 0 cosθ = $$frac{1}{4}$$ Then, 2tanθ + 4tanθ - secθ $$eqalign{
& = 2 imes frac{{sqrt {15} }}{1} + 4 imes frac{{sqrt {15} }}{4} - frac{4}{1} cr
& = 2sqrt {15} + sqrt {15} - 4 cr
& = 3sqrt {15} - 4 cr} $$

[#127] The expression (cos 6 θ + sin 6 θ - 1)(tan 2 θ + cot 2 θ + 2) + 3 is equal to:
Correct Answer

(A) 0

Explanation

Solution: (cos 6 θ + sin 6 θ - 1)(tan 2 θ + cot 2 θ + 2) + 3 Put θ = 45° = (cos 6 45° + sin 6 45° - 1)(tan 2 45° + cot 2 45° + 2) + 3 $$eqalign{
& = left( {frac{1}{8} + frac{1}{8} - 1}
ight)left( {1 + 1 + 2}
ight) + 3 cr
& = - frac{3}{4} imes 4 + 3 cr
& = 0 cr} $$

[#128] If 5sinθ - 4cosθ = 0, 0° < θ < 90°, then the value of $$frac{{5sin heta - 2cos heta }}{{5sin heta + 3cos heta }}$$ xa0 is:
Correct Answer

(C) $$frac{2}{7}$$

Explanation

Solution: $$eqalign{
& 5sin heta - 4cos heta = 0 cr
& 5sin heta = 4cos heta cr
& frac{{sin heta }}{{cos heta }} = frac{4}{5} cr
& an heta = frac{4}{5} cr
& frac{{5sin heta - 2cos heta }}{{5sin heta + 3cos heta }} cr
& = frac{{5 an heta - 2}}{{5 an heta + 3}} cr
& = frac{{5 imes frac{4}{5} - 2}}{{5 imes frac{4}{5} + 2}} cr
& = frac{2}{7} cr} $$

[#129] $${left( {frac{{1 - an heta }}{{1 - cot heta }}}
ight)^2} + 1 = ?$$
Correct Answer

(B) sec 2 θ

Explanation

Solution: $$eqalign{
& {left( {frac{{1 - an heta }}{{1 - cot heta }}}
ight)^2} + 1 cr
& = {left( {frac{{1 - an heta }}{{frac{{ an heta - 1}}{{ an heta }}}}}
ight)^2} + 1 cr
& = frac{{left( {1 - an heta }
ight){{ an }^2} heta }}{{left( {1 - an heta }
ight)}} + 1 cr
& = { an ^2} heta + 1 cr
& = {sec ^2} heta cr} $$

[#130] The value of $${left( {frac{{1 - cot heta }}{{1 - an heta }}}
ight)^2} + 1,$$ xa0 xa00° < θ < 90°, is equal to:
Correct Answer

(A) cosec 2 θ

Explanation

Solution: $$eqalign{
& {left( {frac{{1 - cot heta }}{{1 - an heta }}}
ight)^2} + 1 cr
& = {left( {frac{{1 - frac{{cos heta }}{{sin heta }}}}{{1 + frac{{sin heta }}{{cos heta }}}}}
ight)^2} + 1 cr
& = {left( {frac{{sin heta - cos heta }}{{sin heta }} imes frac{{cos heta }}{{cos heta - sin heta }}}
ight)^2} + 1 cr
& = {left( {frac{{sin heta - cos heta imes cos heta }}{{ - sin heta left( {sin heta - cos heta }
ight)}}}
ight)^2} + 1 cr
& = {left( {frac{{ - cos heta }}{{sin heta }}}
ight)^2} + 1 cr
& = {cot ^2} heta + 1 cr
& = { ext{cose}}{{ ext{c}}^2} heta cr} $$