Trigonometry - Study Mode

[#116] The numerical value of $$frac{5}{{{ ext{se}}{{ ext{c}}^2} heta }}$$ xa0+ $$frac{2}{{1 + { ext{co}}{{ ext{t}}^2} heta }}$$ xa0+ $${ ext{3}}{sin ^2} heta $$ xa0 is?
Correct Answer

(A) 5

Explanation

Solution: $$eqalign{
& frac{5}{{{ ext{se}}{{ ext{c}}^2} heta }}{ ext{ + }}frac{2}{{1 + { ext{co}}{{ ext{t}}^2} heta }}{ ext{ + 3}}{sin ^2} heta cr
& = 5{cos ^2} heta + frac{2}{{{ ext{cose}}{{ ext{c}}^2} heta }} + 3{sin ^2} heta cr
& = 5{ ext{co}}{{ ext{s}}^2} heta + 2{sin ^2} heta + 3{sin ^2} heta cr
& = 5left( {{ ext{co}}{{ ext{s}}^2} heta + {{sin }^2} heta }
ight) cr
& left( {x08ecause {{sin }^2} heta + { ext{co}}{{ ext{s}}^2} heta = 1}
ight) cr
& = 5 imes 1 cr
& = 5 cr} $$

[#117] The numerical value of $$left( {frac{1}{{cos heta }} + frac{1}{{cot heta }}}
ight)$$ xa0 $$left( {frac{1}{{cos heta }} - frac{1}{{cot heta }}}
ight)$$ xa0 is?
Correct Answer

(C) 1

Explanation

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

[#118] The value of cos1° cos2° cos3° ............. cos177° cos178° cos179° is?
Correct Answer

(A) 0

Explanation

Solution: cos1° cos2° cos3° ............. cos177° cos178° cos179° = cos90° = 0[0 will make whole series 0] = 0

[#119] The value of $$frac{{3left( {{{cot }^2}{{47}^ circ } - {{sec }^2}{{43}^ circ }}
ight) - 2left( {{{ an }^2}{{23}^ circ } - { ext{cose}}{{ ext{c}}^2}{{67}^ circ }}
ight)}}{{{ ext{cose}}{{ ext{c}}^2}left( {{{68}^ circ } + heta }
ight) - an left( { heta + {{61}^ circ }}
ight) - {{ an }^2}left( {{{22}^ circ } - heta }
ight) + cot left( {{{29}^ circ } - heta }
ight)}},{ ext{is:}}$$
Correct Answer

(D) -1

Explanation

Solution: $$eqalign{
& frac{{3left( {{{cot }^2}{{47}^ circ } - {{sec }^2}{{43}^ circ }}
ight) - 2left( {{{ an }^2}{{23}^ circ } - { ext{cose}}{{ ext{c}}^2}{{67}^ circ }}
ight)}}{{{ ext{cose}}{{ ext{c}}^2}left( {{{68}^ circ } + heta }
ight) - an left( { heta + {{61}^ circ }}
ight) - {{ an }^2}left( {{{22}^ circ } - heta }
ight) + cot left( {{{29}^ circ } - heta }
ight)}} cr
& = frac{{3left( {{{ an }^2}{{43}^ circ } - {{sec }^2}{{43}^ circ }}
ight) - 2left( {{{ an }^2}{{23}^ circ } - {{sec }^2}{{23}^ circ }}
ight)}}{{{ ext{cose}}{{ ext{c}}^2}left( {{{68}^ circ } + heta }
ight) - an left( { heta + {{61}^ circ }}
ight) - {{cot }^2}left( {{{68}^ circ } - heta }
ight) + an left( {{{61}^ circ } - heta }
ight)}} cr
& = frac{{3 imes left( { - 1}
ight) - 2 imes left( { - 1}
ight)}}{1} cr
& = frac{{ - 3 + 2}}{1} cr
& = - 1 cr} $$

[#120] $$frac{{left( {2sin A}
ight)left( {1 + sin A}
ight)}}{{1 + sin A + cos A}}$$ xa0 xa0is equal to:
Correct Answer

(B) 1 + sinA - cosA

Explanation

Solution: In these type of questions we can go through option- From option B $$eqalign{
& frac{{left( {2sin A}
ight)left( {1 + sin A}
ight)}}{{left( {1 + sin A}
ight) + cos A}} = 1 - cos A + sin A cr
& 2sin Aleft( {1 + sin A}
ight) = {left( {1 + sin A}
ight)^2} - {cos ^2}A cr
& 2sin Aleft( {1 + sin A}
ight) = 1 + sin 2A + 2sin A - 1 + {sin ^2}A cr
& 2sin Aleft( {1 + sin A}
ight) = 2sin Aleft( {1 + sin A}
ight) cr
& { ext{L}}{ ext{.H}}{ ext{.S}}{ ext{.}} = { ext{R}}{ ext{.H}}{ ext{.S}}{ ext{.}} cr
& cr
& {x08f{Alternate:}} cr
& { ext{Put }}A = {90^ circ } cr
& frac{{left( {2sin {{90}^ circ }}
ight)left( {1 + sin {{90}^ circ }}
ight)}}{{1 + sin {{90}^ circ } + cos {{90}^ circ }}} = 2 cr
& { ext{Now from option B}} cr
& 1 + sin A - cos A cr
& = 1 + sin {90^ circ } - cos {90^ circ } cr
& = 1 + 1 - 0 cr
& = 2 cr} $$