Trigonometry - Study Mode
[#171] $$sqrt {frac{{1 + { ext{sin }} heta }}{{1 - { ext{sin }} heta }}} $$ xa0xa0 + $$sqrt {frac{{1 - { ext{sin }} heta }}{{1 + { ext{sin }} heta }}} $$ xa0xa0 is equal to?
Correct Answer
(D) 2secθ
Explanation
Solution: $$eqalign{
& sqrt {frac{{1 + { ext{sin }} heta }}{{1 - { ext{sin }} heta }}} { ext{ + }}sqrt {frac{{1 - { ext{sin }} heta }}{{1 + { ext{sin }} heta }}} cr
& = frac{{{{left( {sqrt {1 + { ext{sin }} heta } }
ight)}^2} + {{left( {sqrt {1 - { ext{sin }} heta } }
ight)}^2}}}{{sqrt {1 - {{sin }^2} heta } }} cr
& = frac{{1 + { ext{sin }} heta + { ext{1}} - { ext{sin }} heta }}{{{ ext{cos}} heta }} cr
& = frac{2}{{{ ext{cos}} heta }} cr
& = 2{ ext{ sec}} heta cr} $$ Alternate shortcut method : $$eqalign{
& { ext{Put }} heta = {30^ circ } cr
& sqrt {frac{{1 + { ext{sin }}30^ circ }}{{1 - { ext{sin }}30^ circ }}} { ext{ + }}sqrt {frac{{1 - { ext{sin }}30^ circ }}{{1 + { ext{sin }}30^ circ }}} cr
& Rightarrow sqrt {frac{{1 + frac{1}{2}}}{{1 - frac{1}{2}}}} + sqrt {frac{{1 - frac{1}{2}}}{{1 + frac{1}{2}}}} cr
& Rightarrow sqrt {frac{3}{1}} + sqrt {frac{1}{3}} cr
& Rightarrow frac{4}{{sqrt 3 }} cr} $$ Now check with option by puting θ = 30° $$eqalign{
& { ext{ = 2 sec 3}}{0^ circ } cr
& = frac{{2 imes 2}}{{sqrt 3 }} cr
& = frac{4}{{sqrt 3 }} cr} $$
[#172] If $${left( {rcos heta - sqrt 3 }
ight)^2}$$ xa0xa0 + $${left( {rsin heta - 1}
ight)^2}$$ xa0 = 0, then the value of $$frac{{r an heta + sec heta }}{{rsec heta + an heta }}$$ xa0 is equal to?
Correct Answer
(A) $$frac{4}{5}$$
Explanation
Solution: $$eqalign{
& {left( {rcos heta - sqrt 3 }
ight)^2} + {left( {rsin heta - 1}
ight)^2} = 0 cr
& Rightarrow {left( {rcos heta - sqrt 3 }
ight)^2} = 0,{ ext{ }}{left( {rsin heta - 1}
ight)^2} = 0 cr
& Rightarrow r{ ext{ cos}} heta = sqrt 3 ........(i) cr
& Rightarrow rsin heta = 1.........(ii) cr
& { ext{Squaring and adding equation (i) and (ii)}} cr
& Rightarrow {r^2}{cos ^2} heta + {r^2}{sin ^2} heta = 3 + 1 cr
& Rightarrow {r^2}left( {{ ext{co}}{{ ext{s}}^2} heta + {{sin }^2} heta }
ight) = 4 cr
& Rightarrow {r^2} = 4 cr
& Rightarrow r = 2 cr
& { ext{tan}} heta = frac{{rsin heta }}{{r{ ext{ cos}} heta }} = frac{1}{{sqrt 3 }}{ ext{ and }}r{ ext{ cos}} heta = sqrt 3 cr
& cos heta = frac{{sqrt 3 }}{r}, cr
& sec heta = frac{r}{{sqrt 3 }} cr
& frac{{r an heta + sec heta }}{{rsec heta + an heta }} cr
& = frac{{frac{r}{{sqrt 3 }} + frac{r}{{sqrt 3 }}}}{{frac{{{r^2}}}{{sqrt 3 }} + frac{1}{{sqrt 3 }}}} cr
& = frac{{rleft( {frac{2}{{sqrt 3 }}}
ight)}}{{frac{{{r^2} + 1}}{{sqrt 3 }}}} cr
& = frac{{2r}}{{{r^2} + 1}} cr
& = frac{{2 imes 2}}{{{2^2} + 1}} cr
& = frac{4}{5} cr
& cr
& {x08f{Alternate:}} cr
& r = 2 cr
& { ext{tan}} heta = frac{{rsin heta }}{{r{ ext{ cos}} heta }} = frac{1}{{sqrt 3 }} cr
& heta = {30^ circ } cr
& = frac{{2 an {{30}^ circ } + sec {{30}^ circ }}}{{2sec {{30}^ circ } + an {{30}^ circ }}} cr
& = frac{{2 imes frac{1}{{sqrt 3 }} + frac{2}{{sqrt 3 }}}}{{2 imes frac{2}{{sqrt 3 }} + frac{1}{{sqrt 3 }}}} cr
& = frac{4}{5} cr} $$
[#173] Let A, B, C, D be the angles of a quadrilateral. If they are concyclic, then the value of cos A + cos B + cos C + cos D is ?
Correct Answer
(A) 0
Explanation
Solution: ∠A + ∠C = ∠B + ∠D = 180° ∴ ∠A = 180° - ∠C cosA = cos(180° - C) ⇒ -cosC Similarly, cosB = -cosD ⇒ cosA + cosB + cosC + cosD ⇒ cosA + cosB - cosA - cosB = 0 Alternate Solution : Put, A = B = C = D = 90° = cosA + cosB + cosC + cosD = cos90° + cos90° + cos90° + cos90° = 0 + 0 + 0 + 0 = 0
[#174] If $$frac{{{ ext{sec}} heta + { ext{tan}} heta }}{{{ ext{sec}} heta - { ext{tan}} heta }} = 2frac{{51}}{{79}}{ ext{,}}$$ xa0 xa0 then the value of $$sin heta $$ xa0is?
Correct Answer
(C) $$frac{{65}}{{144}}$$
Explanation
Solution: $$eqalign{
& { ext{Given, }}frac{{{ ext{sec}} heta + { ext{tan}} heta }}{{{ ext{sec}} heta - { ext{tan}} heta }} = { ext{2}}frac{{51}}{{79}} cr
& Rightarrow frac{{{ ext{sec}} heta + { ext{tan}} heta }}{{{ ext{sec}} heta - { ext{tan}} heta }} = frac{{209}}{{79}} cr
& { ext{By componendo dividendo}} cr
& left[ {frac{a}{b} = frac{c}{d},{ ext{ }}frac{{a + b}}{{a - b}} = frac{{c + d}}{{c - d}}}
ight] cr
& Rightarrow frac{{{ ext{sec}} heta + { ext{tan}} heta + { ext{sec}} heta - { ext{tan}} heta }}{{{ ext{sec}} heta + { ext{tan}} heta - { ext{sec}} heta + { ext{tan}} heta }} = frac{{209 + 79}}{{209 - 79}} cr
& Rightarrow frac{{2sec heta }}{{2{ ext{tan}} heta }} = frac{{288}}{{130}} cr
& Rightarrow frac{{sec heta }}{{{ ext{tan}} heta }} = frac{{288}}{{130}} cr
& Rightarrow frac{{frac{1}{{{ ext{cos}} heta }}}}{{frac{{sin heta }}{{{ ext{cos}} heta }}}} = frac{{288}}{{130}} cr
& Rightarrow frac{1}{{sin heta }} = frac{{288}}{{130}} cr
& Rightarrow { ext{Therefore, }}sin heta = frac{{130}}{{288}} cr
& Rightarrow sin heta = frac{{65}}{{144}} cr} $$
[#175] If 1 + cos 2 θ = 3sinθ cosθ, then the integral value of cotθ is $$left( {0 < heta < frac{pi }{2}}
ight) = ,?$$
Correct Answer
(B) 1
Explanation
Solution: $$eqalign{
& { ext{Given,}} cr
& { ext{1}} + { ext{co}}{{ ext{s}}^2} heta = { ext{3}}sin heta .{ ext{cos}} heta { ext{ }}left( {0 < heta < frac{pi }{2}}
ight) cr
& { ext{1}} + { ext{co}}{{ ext{s}}^2} heta = { ext{3}}sin heta .{ ext{cos}} heta cr
& { ext{Dividing by }}{sin ^2} heta { ext{ in both sides}} cr
& Rightarrow frac{{1 + { ext{co}}{{ ext{s}}^2} heta }}{{{{sin }^2} heta }} = frac{{3sin heta .{ ext{cos}} heta }}{{{{sin }^2} heta }} cr
& Rightarrow { ext{cose}}{{ ext{c}}^2} heta + { ext{co}}{{ ext{t}}^2} heta = 3{ ext{cot}} heta cr
& Rightarrow 1 + { ext{co}}{{ ext{t}}^2} heta + { ext{co}}{{ ext{t}}^2} heta = 3cot heta cr
& left[ {x08ecause 1 + { ext{co}}{{ ext{t}}^2} heta = { ext{cose}}{{ ext{c}}^2} heta }
ight] cr
& Rightarrow 1 + 2{ ext{co}}{{ ext{t}}^2} heta = 3cot heta cr
& Rightarrow 2{ ext{co}}{{ ext{t}}^2} heta = 3cot heta - 1 cr
& { ext{Let }} heta = {45^ circ } cr
& x08ecause { ext{cot4}}{{ ext{5}}^ circ } = 1 cr
& Rightarrow 2{ ext{co}}{{ ext{t}}^2}{45^ circ } - 3{ ext{cot}}{45^ circ } + 1 = 0 cr
& Rightarrow 2 - 3 + 1 = 0 cr
& Rightarrow 0 = 0 cr
& { ext{Therefore cot}} heta = { ext{cot}}{45^ circ } cr
& ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, = 1 cr} $$