Trigonometry - Study Mode

[#1] If P + Q + R = 60°, then what is the value of cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP)?
Correct Answer

(A) $$frac{1}{2}$$

Explanation

Solution: ∵ P + Q + R = 60° By putting P = 0°, Q = 0° and R = 60° ⇒ cosQcosR(cosP - sinP) + sinQsinR(sinP - cosP) ⇒ 1 × cos60°(cos0° - sin0°) + sin0°.sin60°(sin0° - cos0°) ⇒ $$frac{1}{2}$$ (1 - 0) + 0 ⇒ $$frac{1}{2}$$

[#2] The value of $$left[ {frac{{{{sin }^2}{{24}^ circ } + {{sin }^2}{{66}^ circ }}}{{{{cos }^2}{{24}^ circ } + {{cos }^2}{{66}^ circ }}} + {{sin }^2}{{61}^ circ } + cos {{61}^ circ }sin {{29}^ circ }}
ight]$$ xa0 xa0 xa0 xa0is equal to
Correct Answer

(C) 2

Explanation

Solution: $$eqalign{
& left[ {frac{{{{sin }^2}{{24}^ circ } + {{sin }^2}{{66}^ circ }}}{{{{cos }^2}{{24}^ circ } + {{cos }^2}{{66}^ circ }}} + {{sin }^2}{{61}^ circ } + cos {{61}^ circ }sin {{29}^ circ }}
ight] cr
& = frac{1}{1} + {sin ^2}{61^ circ } + cos {61^ circ }sin left( {{{90}^ circ } - {{61}^ circ }}
ight) cr
& = 1 + {sin ^2}{61^ circ } + {cos ^2}{61^ circ } cr
& = 1 + 1 cr
& = 2 cr} $$

[#3] What is $$sin alpha - sin x08eta = ?$$
Correct Answer

(A) $$2cos frac{{alpha + x08eta }}{2}sin frac{{alpha - x08eta }}{2}$$

Explanation

Solution: $$sin alpha - sin x08eta = 2cos frac{{alpha + x08eta }}{2}.sin frac{{alpha - x08eta }}{2}$$

[#4] If 6tanA(tanA + 1) = 5 - tanA, given that 0 < A < $$frac{pi }{2}$$ what is the value of (sinA + cosA)?
Correct Answer

(D) $$frac{3}{{sqrt 5 }}$$

Explanation

Solution: 6tanA(tanA + 1) = 5 - tanA 6tan 2 A + 6tanA + tanA = 5 6tan 2 A + 7tanA = 5 6tan 2 A + 7tanA - 5 = 0 6tan 2 A + 10tanA - 3tanA - 5 = 0 2tanA(3tanA + 5) - 1(3tanA + 5) = 0 (2tanA - 1)(3tanA + 5) = 0 tanA = $$frac{1}{2}$$ and tanA = $$ - frac{5}{2}$$ taking tanA $$ = frac{1}{2} = frac{P}{B}$$ H = √5 $$eqalign{
& herefore sin A + cos A cr
& = frac{P}{H} + frac{B}{H} cr
& = frac{1}{{sqrt 5 }} + frac{2}{{sqrt 5 }} cr
& = frac{3}{{sqrt 5 }} cr} $$

[#5] If tan 2 A - 6tanA + 9 = 0, 0° < A < 90°, What is the value of 6cotA + $$8sqrt {10} $$ cosA?
Correct Answer

(C) 10

Explanation

Solution: tan 2 A - 6tanA + 9 = 0 tan 2 A - 3tanA - 3tanA + 9 = 0 tanA(tanA - 3) - 3(tanA - 3) = 0 (tanA - 3)(tanA - 3) = 0 tanA $$ = frac{3}{1} = frac{P}{B}$$ H 2 = P 2 + B 2 $$eqalign{
& H = sqrt {9 + 1} = sqrt {10} cr
& 6cot A + 8sqrt {10} cos A cr
& = 6left( {frac{B}{P}}
ight) + 8sqrt {10} left( {frac{B}{H}}
ight) cr
& = 6left( {frac{1}{3}}
ight) + 8sqrt {10} left( {frac{1}{{sqrt {10} }}}
ight) cr
& = 2 + 8 cr
& = 10 cr} $$