Trigonometry - Study Mode
[#236] If $$sec heta - an heta = frac{1}{{sqrt 3 }}{ ext{,}}$$ xa0xa0 the value of $$sec heta $$ . $$ an heta $$xa0 = ?
Correct Answer
(C) $$frac{2}{3}$$
Explanation
Solution: $$eqalign{
& {x08f{Shortcut,, method:}} cr
& { ext{Put }} heta = {30^ circ } cr
& Rightarrow sec heta - an heta = frac{1}{{sqrt 3 }} cr
& Rightarrow sec {30^ circ } - an {30^ circ } = frac{1}{{sqrt 3 }} cr
& Rightarrow frac{2}{{sqrt 3 }} - frac{1}{{sqrt 3 }} = frac{1}{{sqrt 3 }} cr
& Rightarrow frac{1}{{sqrt 3 }} = frac{1}{{sqrt 3 }}{ ext{ }}left( {{ ext{Satisfied}}}
ight) cr
& sec heta = {30^ circ } cr
& Rightarrow sec heta . an heta cr
& Rightarrow sec {30^ circ }. an {30^ circ } cr
& Rightarrow frac{2}{{sqrt 3 }} imes frac{1}{{sqrt 3 }} cr
& Rightarrow frac{2}{3} cr} $$
[#237] The value of (cosec a - sin a)(sec a - cos a)(tan a + cot a) is?
Correct Answer
(D) 1
Explanation
Solution: (cosec a - sin a) (sec a - cos a) (tan a + cot a) Put a = 45° ⇒ (cosec45° - sin45°) (sec45° - cos45°) (tan45° + cot45°) $$eqalign{
& Rightarrow left( {sqrt 2 - frac{1}{{sqrt 2 }}}
ight)left( {sqrt 2 - frac{1}{{sqrt 2 }}}
ight)left( {1 + 1}
ight) cr
& Rightarrow frac{1}{{sqrt 2 }} imes frac{1}{{sqrt 2 }} imes 2 cr
& Rightarrow frac{1}{2} imes 2 cr
& Rightarrow 1 cr} $$
[#238] The value of θ(0 ≤ θ ≤ 90°) satisfying 2sin 2 θ = 3cosθ is?
Correct Answer
(A) 60°
Explanation
Solution: $$eqalign{
& { ext{Hit and Trial method}} cr
& { ext{Put }} heta = {60^ circ }{ ext{option A}} cr
& Rightarrow 2{sin ^2}{60^ circ } = 3cos {60^ circ } cr
& Rightarrow 2{left( {frac{{sqrt 3 }}{2}}
ight)^2} = 3left( {frac{1}{2}}
ight) cr
& Rightarrow frac{3}{2} = frac{3}{2}{ ext{ }}left( {{ ext{LHS}} = { ext{RHS}}}
ight) cr} $$
[#239] a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a 2 + b 2 + c 2 = ab + bc + ca, then the value of (sin 2 A + sin 2 B + sin 2 C) is?
Correct Answer
(D) $$frac{9}{4}$$
Explanation
Solution: $$eqalign{
& {a^2} + {b^2} + {c^2} = ab + bc + ca cr
& Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 cr
& Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0 cr
& Rightarrow {a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca = 0 cr
& Rightarrow {left( {a - b}
ight)^2} + {left( {b - c}
ight)^2} + {left( {c - a}
ight)^2} = 0 cr
& herefore a = b = c cr
& vartriangle { ext{ABC}} = { ext{equilateral }}vartriangle cr
& herefore angle { ext{A}} = angle { ext{B}} = angle { ext{C}} = {60^ circ } cr
& { ext{So, }}{sin ^2}{ ext{A}} + {sin ^2}{ ext{B}} + {sin ^2}{ ext{C}} cr
& Rightarrow {sin ^2}{60^ circ } + {sin ^2}{60^ circ } + {sin ^2}{60^ circ } cr
& Rightarrow 3{sin ^2}{60^ circ } cr
& Rightarrow 3 imes {left( {frac{{sqrt 3 }}{2}}
ight)^2} cr
& Rightarrow frac{9}{4} cr} $$
[#240] If $${ ext{tan }}alpha = 2,$$ xa0 then the value of $$frac{{{ ext{cose}}{{ ext{c}}^2}alpha - { ext{se}}{{ ext{c}}^2}alpha }}{{{ ext{cose}}{{ ext{c}}^2}alpha + se{c^2}alpha }}$$ xa0 is?
Correct Answer
(C) $$ - frac{3}{5}$$
Explanation
Solution: $$eqalign{
& { ext{tan}}alpha = 2left( {{ ext{given}}}
ight) cr
& herefore frac{{{ ext{cose}}{{ ext{c}}^2}alpha - { ext{se}}{{ ext{c}}^2}alpha }}{{{ ext{cose}}{{ ext{c}}^2}alpha + se{c^2}alpha }} cr} $$ (Divide by coses 2 α both in N and D) $$eqalign{
& = frac{{1 - { ext{ta}}{{ ext{n}}^2}alpha }}{{1 + { ext{ta}}{{ ext{n}}^2}alpha }} cr
& = frac{{1 - {{left( 2
ight)}^2}}}{{1 + {{left( 2
ight)}^2}}} cr
& = - frac{3}{5} cr} $$