Trigonometry - Study Mode

[#111] The simplified value of (secA - cosA) 2 + (cosecA - sinA) 2 - (cotA - tanA) 2
Correct Answer

(C) 1

Explanation

Solution: (secA - cosA) 2 + (cosecA - sinA) 2 - (cotA - tanA) 2 = (sec 2 A + cos 2 A - 2secA.cosA) + (coses 2 A + sin 2 A - 2cosecA.sinA) - (cot 2 A + tan 2 A - 2cotA.tanA) = sec 2 A - tan 2 A + cos 2 A + sin 2 A + coses 2 A - cot 2 A - 2 = 3 - 2 = 1 Alternate shortcut method: (secA - cosA) 2 + (cosecA - sinA) 2 - (cotA - tanA) 2 Put θ = 45° = (sec45° - cos45°) 2 + (cosec45° - sin45°) 2 - (cot45° - tan45°) 2 $$eqalign{
& = {left( {sqrt 2 - frac{1}{{sqrt 2 }}}
ight)^2} + {left( {sqrt 2 - frac{1}{{sqrt 2 }}}
ight)^2} - {left( {1 - 1}
ight)^2} cr
& = frac{1}{2} + frac{1}{2} - 0 cr
& = 1 cr} $$

[#112] The angles of a triangle are (x + 5)°, (2x - 3)° and (3x + 4)°. Then the value of x is?
Correct Answer

(C) 29

Explanation

Solution: (x + 5)° + (2x - 3)° + (3x + 4)° = 180° (Sum of all angles in triangle is 180°) ⇒ 6x + 6° = 180° ⇒ (x + 1) = 30° ⇒ x = 29°

[#113] The expression $$frac{{ an {{57}^ circ } + cot {{37}^ circ }}}{{ an {{33}^ circ } + cot {{53}^ circ }}}$$ xa0xa0 is equal to?
Correct Answer

(B) tan57° cot37°

Explanation

Solution: $$eqalign{
& frac{{ an {{57}^ circ } + cot {{37}^ circ }}}{{ an {{33}^ circ } + cot {{53}^ circ }}} cr
& = frac{{cot{{33}^ circ } + an {{53}^ circ }}}{{ an {{33}^ circ } + cot {{53}^ circ }}} cr
& = frac{{frac{1}{{ an {{33}^ circ }}} + an {{53}^ circ }}}{{ an {{33}^ circ } + frac{1}{{ an {{53}^ circ }}}}} cr
& = frac{{1 + an {{53}^ circ }. an {{33}^ circ }}}{{ an {{33}^ circ }. an {{53}^ circ } + 1}} imes frac{{ an {{53}^ circ }}}{{ an {{33}^ circ }}} cr
& = an {53^ circ }.cot{33^ circ } cr
& = cot {37^ circ }. an {57^ circ } cr} $$

[#114] The value of $$frac{{cot {{30}^ circ } - cot {{75}^ circ }}}{{ an {{15}^ circ } - an {{60}^ circ }}}$$ xa0xa0 is?
Correct Answer

(D) -1

Explanation

Solution: $$eqalign{
& frac{{cot {{30}^ circ } - cot {{75}^ circ }}}{{ an {{15}^ circ } - an {{60}^ circ }}} cr
& = frac{{ an {{60}^ circ } - an {{15}^ circ }}}{{ an {{15}^ circ } - an {{60}^ circ }}} cr
& = frac{{ - left( { an {{15}^ circ } - an {{60}^ circ }}
ight)}}{{ an {{15}^ circ } - an {{60}^ circ }}} cr
& = - 1 cr} $$

[#115] If $${ ext{se}}{{ ext{c}}^2} heta + { ext{ta}}{{ ext{n}}^2} heta = frac{7}{{12}}{ ext{,}}$$ xa0 xa0 then $${ ext{se}}{{ ext{c}}^4} heta $$ xa0- $${ ext{ta}}{{ ext{n}}^4} heta $$ xa0 = ?
Correct Answer

(A) $$frac{7}{{12}}$$

Explanation

Solution: $$eqalign{
& left( {{ ext{se}}{{ ext{c}}^4} heta - { ext{ta}}{{ ext{n}}^4} heta }
ight) cr
& Rightarrow left( {{ ext{se}}{{ ext{c}}^2} heta - { ext{ta}}{{ ext{n}}^2} heta }
ight)left( {{ ext{se}}{{ ext{c}}^2} heta + { ext{ta}}{{ ext{n}}^2} heta }
ight) cr
& Rightarrow 1 imes left( {{ ext{se}}{{ ext{c}}^2} heta + { ext{ta}}{{ ext{n}}^2} heta }
ight)[1 + { ext{ta}}{{ ext{n}}^2} heta = { ext{se}}{{ ext{c}}^2} heta ] cr
& Rightarrow 1 imes frac{7}{{12}} cr
& Rightarrow frac{7}{{12}} cr} $$