Solution/Answer :
sin A = \(3 \over 5\)

In ∆ABC, ∠B = 90°

AC = 5 and BC = 3


∴ AB = \( \sqrt{AC^2 - BC^2} \) = \( \sqrt{5^2 - 3^2} \)

= \(\sqrt{25 - 9}\) = \(\sqrt{16}\) = 4

Now, cos θ = \(AB \over AC\) = \(4 \over 5\)

tanθ = \(BC \over AB\) = \(3 \over 4\)

cotθ = \(1 \over tan θ\) = \(4 \over 3\)

sec θ = \(1 \over cos θ\) = \(5 \over 4\)

cosec θ = \(1 \over sin θ\) = \(5 \over 3\)

Solution/Answer :
sec A = \(17 \over 8\) (A is an acute angle)

In right ∆ABC

sec A = \(AC \over AB\) = \(17 \over 8\)

AC = 17, AB = 8



BC = \( \sqrt{AC^2 - AB^2} \) = \( \sqrt{17^2 - 8^2} \)

= \(\sqrt{289 - 64}\) = \(\sqrt{225}\) = 15

Now, sin A = \(BC \over AC\) = \(15 \over 17\)

cos A = \(1 \over sec A \) = \(8 \over 17\)

tan A = \(BC \over AB\) = \( 15 \over 8 \)

cot A = \(1 \over tan A \) = \(8 \over 15\)

cosec A = \(1 \over sin A \) = \(17 \over 15\)

Solution/Answer :

Given,

12 cosec θ = 13

⇒ cosec θ = 13/12

In right ∆ ABC,

∠A = θ

So, cosec θ = AC/BC = 13/12

AC = 13 and BC = 12

By Pythagoras theorem,

AB = √(AC2 – BC2)

= √[(13)2 – (12)2]

= √(169 – 144)

= √25

= 5

Now,

sin θ = BC/AC = 12/13

cos θ = AB/AC = 5/13

Hence,

\( {2sin θ - 3cos θ} \over {4sin θ - 9cos θ} \) = \( {2 x { 12 \over 13 } - 3 x { 5 \over 13 } } \over {4 x { 12 \over 13 } - 9 x { 5 \over 13 } } \)

= \( { 24 \over 13 } - { 15 \over 13 } \over { 48 \over 13 } - { 45 \over 13 } \) = \( { 9 \over 13 } \over { 3 \over 13 } \)

= \( { 9 \over 13 } x { 13 \over 3 } \) = 3

Solution/Answer :
Given that
(i) \(cos^2 26° + cos 64° sin 26° + { tan 36° \over cot 54° } \)

= \(cos^2 26° + cos (90° - 26°) sin 26° + { tan 36° \over cot (90° - 36°) } \)

= \(cos^2 26° + sin^2 26° + { tan 36° \over tan 36° } \)

= 1 + 1 = 2

[∵ cos(90° - θ) = sin θ, cot (90° - θ) = tan θ and \( {sin^2} θ + {cos^2} θ = 1 \) ]

(ii) \( { sec 17° \over cosec 73° } + {tan68° \over cot 22° } + cos^2 44° + cos^2 46° \)

= \( { sec (90° - 73°) \over cosec 73° } + {tan (90° - 22°) \over cot 22° } + cos^2 (90° - 46°) + cos^2 46° \)

= \( { cosec 73° \over cosec 73° } + { cot 22° \over cot 22° } + sin^2 46° + cos^2 46° \)

= 1 + 1 + 1 = 3

Solution/Answer :
given that (i) \( {sin 65° \over cos 25°} + { cos 32° \over sin 58° } – sin 28° x sec 62° + cosec² 30° (2015) \)

= \( {sin 65° \over cos (90° - 65°)} + { cos 32° \over sin (90°-32°) } – sin 28° x sec (90° - 28°) + (2)^2 \)

= \( {sin 65° \over sin 65°} + { cos 32° \over cos 32° } – sin 28° x cosec 28° + (2)^2 \)

= \( 1 + 1 – 1 + 4 \)

= 6 - 1 = 5

(ii) \({sec29° \over cosec61°} + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3(sin^2 38° + sin^2 52°). \)

= \({sec29° \over cosec(90° - 29°)} + 2 cot 8° cot 17° cot 45° cot (90° - 17°) cot (90°-8°) – 3(sin^2 38° + sin^2 (90° - 38°) \)

= \({sec29° \over cosec(90° - 29°)} + 2 cot 8° cot 17° cot 45° cot (90° - 17°) cot (90°-8°) – 3(sin^2 38° + sin^2 (90° - 38°) \)

= \({sec29° \over cosec(90° - 29°)} + 2 cot 8° cot 17° cot 45° cot (90° - 17°) cot (90°-8°) – 3(sin^2 38° + sin^2 (90° - 38°) \)

= \({sec29° \over sec29°} + 2 cot 8° cot 17° cot 45° tan 17° tan 8° – 3(sin^2 38° + cos^2 38°) \)

= \( 1 + 2 x cot 8° x cot 17° x cot 45° x {1 \over cot 8°} x {1 \over cot 17°} – {3 x 1} \)

= \( 1 + 2 x 1 x 1 x 1 – 3 \)

= \( 3 – 3 \) = \( 0 \)

Solution/Answer :
Given that

(i) \( {sin35° cos55° + cos35° sin55°} \over { cosec^2 10° − tan^2 80°} \)

= \( {sin35° cos(90° - 35)° + cos35° sin(90° - 35°)} \over { cosec^2 10° − tan^2 (90° - 10°)} \)

= \( {sin35° sin35° + cos35° cos35°} \over { cosec^2 10° − cot^2 10} \)

= \( {sin^2 35° + cos^2 35°} \over { cosec^2 10° − cot^2 10} \)

= \( 1 \over 1 \) = 1

[∵ \( {sin^2} θ + {cos^2} θ = 1 \) and \( {cosec^2} θ - {cot^2} θ = 1 \) ]

(ii) \(sin^2 34° + sin^2 56° + 2tan18° tan72° − cot^2 30° \)

= \(sin^2 34° + sin^2 (90° - 34°) + 2tan18° tan(90° - 18°) − cot^2 30° \)

= \(sin^2 34° + cos^2 34° + 2 tan18° cot18° − cot^2 30° \)

= \(1 + 2 x 1 − (\sqrt 3)^2 \)

= \(1 + 2 − 3 \) = \( 3 − 3 \) = \( 0 \)

Solution/Answer :
(i) \( ({ tan25° \over cosec65° })^2 + ({cot25° \over sec65° })^2 + 2 tan18° tan45° tan72° \)

= \( ({ tan25° \over cosec(90° -25°) })^2 + ({cot25° \over sec(90°-25°) })^2 + 2 tan18° tan45° tan(90° - 18°) \)

= \( ({ tan25° \over sec25° })^2 + ({cot25° \over cosec25° })^2 + 2 tan18° tan45° cot18° \)

= \( ({ {sin25° x cos25°} \over {cos25° x 1} })^2 + ({ {cos25° x sin25°} \over {sin25° x 1} })^2 + 2 x 1 x 1 \)

= \( sin^2 25° + cos^2 25° + 2 \) = \( 1 + 2 \) = \( 3 \)

[∵ \( {sin^2} θ + {cos^2} θ = 1 \) and \( {tan} θ \ {cot} θ = 1 \) ]

(ii) \( {(cos^2 25° + cos^2 65°)} + cosecθ \ sec(90°−θ) − cotθ \ tan(90°−θ) \)

= \( {[ cos^2 25° + cos^2 (90° - 25°) ]} + cosecθ \ cosecθ − cotθ \ cotθ \)

= \( {( cos^2 25° + sin^2 25° )} + cosec^2 θ − cot^2 θ \)

= \( 1 + 1 \) = \( 2 \)

[∵ \( {sin^2} θ + {cos^2} θ = 1 \) and \( cosec^2 θ - cot^2 θ = 1 \) ]

Solution/Answer :
(i) (sec A + tan A) (1 – sin A) = cos A

L.H.S = \( ({1 \over cos A} + { sin A \over cos A}) {( 1 - sin A )} \)

= \( {(1 + sin A) \times (1 - sin A)} \over cos A \)

= \( {cos^2 A} \over cos A \) = \( cos A \) = \( R.H.S \)
[∵ \( 1 - {sin^2} θ = cos^2 θ \) ]
(ii) \( {(1 + tan^2 A) (1 – sin A) (1 + sin A)} \) = 1

L.H.S. = \( {(1 + tan^2 A) (1 – sin A) (1 + sin A)} \)

= \( (1 + {sin^2 A \over cos^2 A}) (1 – sin^2 A) \)

= \( {{cos^2 A + sin^2 A} \over cos^2 A} \times cos^2 A \)

= \( {1 \over cos^2 A} \times cos^2 A \) = \( 1 \) = \( R.H.S \)
[∵ \( 1 - {sin^2} A = cos^2 A \) and \( sin^2 A + cos^2 A = 1 \) ]

Solution/Answer :
(i) tan A + cot A = sec A cosec A

L.H.S = tan A + cot A

= \( {sinA \over cosA} + {cosA \over sinA} \)

= \( {sin^2 A + cos^2 A} \over {sinA \ cosA} \)

= \( 1 \over {sinA \ cosA} \) = \( cosecA \ secA \) = \( secA \ cosecA \) = \( R.H.S. \)
[∵ \( sin^2 A + cos^2 A = 1 \) ]
(ii) (1 – cos A)(1 + sec A) = tan A sin A

L.H.S. = \( (1 – cos A)(1 + sec A) \)

= \( (1 – cos A)(1 + {1 \over cos A}) \)

= \( (1 – cos A){(cosA + 1) \over cos A} \)

= \( {(1 – cos A) (1 + cosA)} \over cos A \)

= \( {1 – cos^2 A} \over cos A \)

= \( sin^2 A \over cos A \) = \( sin A \times {sinA \over cos A} \)

= \( sinA tanA \) = \( tanA sinA \) = \( R.H.S \)
[∵ \( 1 – cos^2 A = sin^2 A \) and \( tanA = {sinA \over cosA} \) ]

Solution/Answer :
(i) \( cot^2 A - cos^2 A = cot^2 A cos^2 A \)
LHS = \( cot^2 A - cos^2 A \)

= \( {cos^2 A \over sin^2A} - cos^2 A \)

= \( {cos^2 A {(1-sin^2A)}}\over sin^2A \)

= \( {cos^2 A {cos^2 A\over sin^2A}} \)

= \( cos^2 A cot^2A = RHS \)

(ii) \( 1 + { tan^2 A \over {1+secA} } = sec A \)

LHS = \( 1 + { tan^2 A \over {1+secA} } \)

= \( 1 + { {sec^2 A -1} \over {1+secA} } \)

= \( 1 + { {{(secA-1)}{(secA+1)}} \over {1+secA} } \)

= \( 1 + { {{(secA-1)}} } \)

= \(secA = RHS\)

(iii) \( {{1+secA} \over secA} = {sin^2 A \over {1-cosA}} \)

LHS = \( {{1+secA} \over secA} \)

= \( {{1+{1 \over cosA}} \over {1 \over cosA}} \)

= \( {{1+{1 \over cosA}} \over {1 \over cosA}} \)

= \( {({1+{1 \over cosA})} \times {cosA}} \)

= \( {{cosA+1} \over cosA} \times {cosA} \)

= \( {cosA +1} \)

= \( {(1+cosA)} \times {{(1-cosA)}\over{(1-cosA)}} \)

= \( {{(1-cos^2A)}\over{(1-cosA)}} \)

= \( {{sin^2A}\over{1-cosA}} \) = RHS

(iv) \( {sinA \over (1-cosA)} = cosec A + cotA \)

LHS = \( sinA \over (1-cosA) \)

= \( {sinA \over (1-cosA)} \times {(1+cosA)\over(1+cosA)} \)

= \( {{sinA + sinA cosA} \over (1-cos^2A)} \)

= \( {{sinA + sinA cosA} \over sin^2A} \)

= \( {sinA \over sin^2A} + {{sinAcosA} \over \sin^2A } \)

= \( {1 \over sinA} + {{cosA} \over sinA } \)

= \( {cosecA} + {cotA } = RHS \)

Solution/Answer :
(i) \( {sinA \over {1+cosA}} = {{1-cosA} \over sinA} \)

LHS = \( {sinA \over {1+cosA}} \)

= \( {sinA \over {1+cosA}} \times {(1-cosA) \over (1-cosA)} \)

= \( {sinA (1-cosA)} \over {(1-cos^2A)} \)

= \( {sinA (1-cosA)} \over {sin^2A} \)

= \( {1-cosA} \over {sinA} \)

(ii) \( {{1-tan^2 A} \over {cot^2 A -1}} = tan^2A \)

LHS = \( {{1-tan^2 A} \over {cot^2 A -1}} \)

= \( {{1-{({sin^2A \over cos^2A})}} \over {{({cos^2A \over sin^2A})} -1}} \)

= \( {{cos^2A - sin^2A} \over cos^2A} \over {{cos^2A - sin^2A} \over sin^2A} \)

= \( {{cos^2A - sin^2A} \over cos^2A} \times { sin^2A \over {cos^2A - sin^2A}} \)

= \( {sin^2A \over cos^2A} = tan^2A = RHS\)


(iii) \( {sinA \over {1+cosA}} = cosecA-cotA \)

LHS = \( {sinA \over {1+cosA}} \)

= \( {sinA \over {1+cosA}} \times {{1-cosA}\over{1-cosA}} \)

= \( {{sinA{(1-cosA)}} \over {1-cos^2A}} \)

= \( {{sinA{(1-cosA)}} \over sin^2A} \)

= \( {{{(1-cosA)}} \over sinA} \)

= \( {1\over sinA} - {cosA \over sinA} \)

= \( {cosecA - cotA} = RHS \)

Solution/Answer :
(i) \( {{secA-1} \over {secA+1}} = {{1-cosA} \over {1+cosA}} \)

LHS = \( {{secA-1} \over {secA+1}} \)

= \( {{{1 \over cosA}-1} \over {{1 \over cosA}+1}} \)

= \( {{1-cosA}\over cosA} \over {{1+cosA}\over cosA} \)

= \( {{1-cosA} \over cosA} \times {{1+cosA}\over cosA} \)

= \( {{1-cosA} \over {1+cosA}} = RHS \)

(ii) \( {tan^2θ \over {(secθ-1)^2}} = {{1+cosθ}\over{1-cosθ}} \)

LHS = \( {tan^2θ \over {(secθ-1)^2}} \)

= \( {sec^2θ -1} \over {(secθ-1)^2} \)

= \( {(secθ+1) (secθ-1)} \over {(secθ-1)^2} \)

= \( {secθ+1}\over {secθ-1} \)

= \( {{1 \over cosθ} +1} \over {{1 \over cosθ} -1} \)

= \( {{1+cosθ}\over cosθ} \over {{1-cosθ} \over cosθ} \)

= \( {{1+cosθ}\over cosθ} \times {cosθ \over {1-cosθ}}\) = \( {{1+cosθ}\over {1-cosθ}} =RHS \)


(iii) \( {(1+tanA)^2} + {(1-tanA)^2} = 2 sec^2 A \)

LHS = \( {(1+tanA)^2} + {(1-tanA)^2} \)

= \( 1+tan²A+2tanA+1+tan²A-2tanA \)

= \( 2+2tan²A \)

from formula, \( sen²A-tan²A=1 \) that is \( sec²A=1+tan²A \)

= \( 2(1+tan²A) \)

= \( 2sec²A = RHS \)

(iv) \( sec^2A +cosec^2A = sec^2A cosec^2A \)

LHS = \( sec^2A +cosec^2A \)

= \( {1 \over cos^2A} + {1 \over sin^2A} \)

= \( {sin^2A + cos^2A} \over {cos^2A sin^2A} \)

= \( 1 \over {cos^2A sin^2A} \)

= \( {1 \over {cos^2A}}\times{1 \over {sin^2A}} \)

= \( sec^2A cosec^2A = RHS \)

Solution/Answer :
(i) \( {{1+sinA} \over cosA} + {cosA \over {1+sinA}} = 2 secA \)

LHS = \( {{1+sinA} \over cosA} + {cosA \over {1+sinA}} \)

= \( {{(1+sinA)^2} + {cos^2A}} \over cosA {(1+sinA)} \)

= \( {{1 + sin^2A + 2sinA + cos^2A} \over cosA {(1+sinA)} } \)

= \( {1+1+2sinA} \over cosA {(1+sinA)} \)

= \( {2+2sinA} \over cosA {(1+sinA)} \)

= \( {2(1+sinA)} \over cosA {(1+sinA)} \)

= \( 2 \over cosA \) = \( 2 secA = RHS \)

(ii) \( {tanA \over {secA-1}} + {tanA \over {secA+1}} = 2 cosecA \)

LHS = \( {tanA \over {secA-1}} + {tanA \over {secA+1}} \)

= \( {{sinA \over cosA} \over {{1 \over cosA} -1}} + {{sinA \over cosA} \over {{1 \over cosA} +1}} \)

= \( {{sinA \over cosA} \over {{1-cosA}\over cosA}} + {{sinA \over cosA} \over {{1+cosA}\over cosA}} \)

= \( {({sinA \over cosA} \times {cosA \over {1-cosA}})} + {({sinA \over cosA} \times {cosA \over {1+cosA}})} \)

= \( {sinA \over {1-cosA}} + {sinA \over {1+cosA}} \)

= \( {{sinA (1+cosA)} + {sinA (1-cosA)} } \over {(1-cosA)(1+cosA)} \)

= \( {{sinA + sinAcosA + sinA - sinAcosA}\over {1-cos^2A}} \)

= \( {2sinA} \over sin^2A \) = \( 2 \over sinA \) = \( 2 cosecA = RHS \)

Solution/Answer :
(i) \( {cosecA \over {cosecA-1}} + {cosecA \over {cosecA+1}} = 2sec^2A \)

LHS = \( {cosecA \over {cosecA-1}} + {cosecA \over {cosecA+1}} \)

= \( {{cosecA (cosecA+1)} + {cosecA (cosecA-1)}} \over {(cosecA-1)(cosecA+1)} \)

= \( {{cosec^2A + cosecA + cosec^2A - cosecA} \over {cosec^2A-1}} \)

= \( {2cosec^2A} \over cot^2A \)

= \( {2 \over sin^2A} \over {cos^2A \over sin^2A} \)

= \( {2 \over sin^2A} \times {sin^2A \over cos^2A} \)

= \( {2 \over cos^2A } = {2 sec^2A} = RHS \)

(ii) \( {cotA-tanA} = {{2cos^2A-1}\over{sinA cosA}} \)

LHS = \( {cosA \over sinA} - {sinA \over cosA} \)

= \( {cos^2A - sin^2A} \over {sinA cosA} \)

= \( {cos^2A - (1-cos^2A)} \over {sinA cosA} \)

= \( {cos^2A -1 + cos^2A} \over {sinA cosA} \)

= \( {{2cos^2A-1} \over {sinA cosA}} = RHS \)

(iii) \( {{cotA-1} \over {2-sec^2A}} = {cotA \over {1+tanA}} \)

LHS = \( {cotA - 1} \over {2 - sec^2A} \)

= \({{cosA \over sinA} - 1} \over {1 + 1 - sec^2A} \)

= \({{cosA - sinA} \over sinA} \over {1 - tan^2A} \)

= \({{cosA - sinA} \over sinA} \over {1 - {sin^2A \over cos^2A}} \)

= \({{cosA - sinA} \over sinA} \over {{cos^2A - sin^2A} \over cos^2A} \)

= \({{cosA - sinA} \over sinA } \times {cos^2A \over {(cosA + sinA)} {(cosA - sinA)}} \)

= \( {cos^2A} \over {sinA (cosA + sinA)} \)

RHS = \( cotA \over {1 - tanA} \)

= \( {(cosA \over sinA)} \over {(1 - sinA) \over cosA} \)

= \( {cosA \over sinA} \over {{cosA - sinA} \over cosA} \)

= \( {cosA \over sinA} \times {cosA \over {cosA - sinA}} \)

= \( {cos^2A} \over {sinA (cosA + sinA)} \)

\( LHS = RHS \)

Hence proved.

Solution/Answer :
(i) \( tan^2θ - sin^2θ = tan^2θsin^2θ \)

LHS = \( tan^2θ - sin^2θ \)

= \( {sin^2θ \over cos^2θ} - sin^2θ \)

= \( {sin^2θ - sin^2θcos^2θ} \over cos^2θ \)

=\( {sin^2θ{(1-cos^2θ)}} \over cos^2θ \)

= \( {{sin^2θ sin^2θ} \over cos^2θ} \)

= \( {sin^2θ \over cos^2θ} sin^2θ \)

= \( tan^2θsin^2θ = RHS \)

(ii) \( {cosθ \over {1-tanθ}} - {sin^2θ \over {cosθ-sinθ}} = cosθ+sinθ \)

LHS = \( {cosθ \over {1-tanθ}} - {sin^2θ \over {cosθ-sinθ}} \)

= \( {cosθ \over {1-{sinθ \over cosθ}}} - {sin^2θ \over {cosθ-sinθ}} \)

= \( {cosθ \over {{cosθ-sinθ} \over cosθ}} - {sin^2θ \over {cosθ-sinθ}} \)

= \( {cosθ \times {cosθ \over {cosθ-sinθ}}} - {sin^2θ \over {cosθ-sinθ}} \)

= \( {cos^2θ \over {cosθ-sinθ}} - {sin^2θ \over {cosθ-sinθ}} \)

= \( {cos^2θ -sin^2θ } \over { cosθ-sinθ} \)

= \( {(cosθ -sinθ)(cosθ +sinθ)} \over {(cosθ-sinθ)} \)

= \( {(cosθ +sinθ)} = RHS \)

Solution/Answer :
(i) \( cosec^4θ-cosec^2θ = cot^4θ+cot^2θ \)

LHS = \( cosec^4θ-cosec^2θ \)

= \( {cosec^2θ(cosec^2θ-1)} \)

= \( cosec^2θ cot^2θ \)

= \( {(1+cot^2θ) cot^2θ} \)

= \( {cot^2θ+} \)

= \( {cot^4θ + cot^2θ} = RHS \)

(ii) \( 2sec^2θ-sec^4θ-2cosec^2θ+cosec^4θ = cot^4θ - tan^4θ \)

LHS = \( 2sec^2θ−sec^4θ−2cosec^2θ+cosec^4θ \)

= \( {(cosec^4θ−2cosec^2θ)}−{(sec^4θ−2sec^2θ)} \)

= \( {(cosec^4θ−2cosec^2θ+1)}−{(sec^4θ−2sec^2θ+1)} \)

= \( {(cosec^2θ−1)^2}−{(sec^2θ−1)^2} \)

= \( {cot^4θ−tan^4θ} \)

= \( RHS \)

Solution/Answer :
(i) \( {{1+cosθ-sin^2θ} \over {sinθ{(1+cosθ)}}} = cotθ \)

LHS = \( {{1+cosθ-sin^2θ} \over {sinθ{(1+cosθ)}}} \)

= \( {{1+cosθ-{1+cos^2θ}} \over {sinθ{(1+cosθ)}}} \)

= \( {{1+cosθ-1+cos^2θ}} \over {sinθ{(1+cosθ)}} \)

= \( {cosθ(1+cosθ)} \over {sinθ{(1+cosθ)}} \)

= \( {cosθ \over sinθ} = cotθ = RHS \)

(ii) \( {{tan^3θ-1}\over{tanθ-1}} = sec^2θ+tanθ \)

LHS = \( {{tan^3θ-1}\over{tanθ-1}} \)

= \( {{(tanθ-1)}{tan^2θ + tanθ +1}} \over {tanθ-1} \)

= \( {tan^2θ + tanθ +1} \)

= \( {(1+ tan^2θ)} + tanθ \)

= \( sec^2θ + tanθ = RHS \)

Solution/Answer :
(i) \( {{1+cosecA} \over cosecA} = {cos^2A \over {1-sinA}} \)

LHS = \( {{1+cosecA} \over cosecA} \)

= \( {1 \over cosecA} + {cosecA \over cosecA} \)

= \( sinA+1 \)

= \( {sinA+1} \times {{(1-sinA )} \over {(1-sinA)}} \)

= \( {1- sin^2A} \over {1-sinA} \)

= \( cos^2A \over {1-sinA} \)

= \( RHS \)

(ii) \( {\sqrt {{1-cosA}\over{1+cosA}}} = {sinA \over {1+cosA}} \)

LHS = \( {\sqrt {{1-cosA}\over{1+cosA}}} \)

= \( {\sqrt {{1-cosA}\over{1+cosA}} \times {{1-cosA}\over{1-cosA}}} \)

= \( {\sqrt {{(1-cosA)^2}\over{1-cos^2A}}} \)

= \( {{(1-cosA)}\over{sinA}} \times {{1+cosA}\over{1+cosA}} \)

= \( {{1-cos^2A}\over {sinA (1+cosA)}} \)

= \( sin^2A \over {sinA (1+cosA)} \)

= \( {sinA \over {1+cosA}} = RHS \)

Solution/Answer :
(i) \( {\sqrt{{1+sinA} \over {1-sinA}}} = tanA + secA \)
LHS= \( {\sqrt{{1+sinA} \over {1-sinA}}} \)

= \( {\sqrt{{1+sinA} \over {1-sinA}} \times {{1+sinA} \over {1+sinA}} } \)

= \( {\sqrt{{(1+sinA)^2} \over {1-sin^2A}} } \)

= \( {\sqrt{{(1+sinA)^2} \over {cos^2A}} } \)

= \( {{{(1+sinA)} \over {cosA}} } \)

= \( {sinA \over cosA} + {1\over cosA} \)

= \( {tanA + secA} = RHS \)

(ii) \( {\sqrt{{1-cosA} \over {1+cosA}}} = cosecA-cotA \)

LHS= \( {\sqrt{{{1-cosA} \over {1+cosA}} \times {{1-cosA} \over {1-cosA}}}} \)

= \( {\sqrt{{(1-cosA)^2} \over {1-cos^2A}}} \)

= \( {\sqrt{{(1-cosA)^2} \over {sin^2A}}} \)

= \( {(1-cosA)} \over {sinA} \)

= \( {1 \over sinA}-{cosA \over sinA} \)

= \( cosecA-cotA = RHS \)

Solution/Answer :
(i) \( {\sqrt{{secA-1}\over{secA+1}}} + {\sqrt{{secA+1}\over{secA-1}}} = 2cosecA \)



(ii) \( {cosAcotA \over {1-sinA}} = 1+ cosecA \)

Solution/Answer :
(i) \( {(1 + tanA) \over sinA} + {(1 + cotA) \over cosA} \) = \( {2(secA + cosecA)} \)

(ii) \( {sec^4A - tan^4A} = {1 + 2 tan^2A} \)

Solution/Answer :
(i) \( {cosec^6A - cot^6A} = {3cot^2A cosec^2A +1} \)

(ii) \( {sec^6A - tan^6A} = {1 + 3tan^2A + 3 tan^4A} \)
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Solution/Answer :
(i) \( {cotθ + cosecθ - 1} \over {cotθ - cosecθ + 1} \) = \( {1 + cosθ} \over sinθ \)


(ii) \( {sinθ} \over {cotθ + cosecθ} \) = \( 2 + {sinθ \over {cotθ - cosecθ}} \)

Solution/Answer :
(i) \( {(sinθ + cosθ)} {(secθ + cosecθ)} = {2 + secθ cosecθ} \)

(ii) \( {(cosecA - sinA)} {(secA - cosA)} sec^2A = tanA \)

Solution/Answer :
(i) \( {{sin^3 + cos^3A} \over {sinA + cosA}} + {{sin^3A - cos^3A} \over {sinA - cosA}} \) = 2

(ii) \( {tan^2A \over {1 + tan^2A}} + {cot^2A \over {1 + cot^2A}} \) = 1

Solution/Answer :
(i) \( {1 \over {secA + tanA}} - {1 \over cosA} = {1 \over cosA} - {1 \over {secA - tanA}} \)

(ii) \( {(sinA + secA)^2} + {(cosA + cosecA)^2} = {(1 + secA cosecA)^2} \)

(iii) \( {tanA + sinA} \over {tanA - sinA} \) = \( {secA + 1} \over {secA - 1} \)

Solution/Answer :
sinθ + cosθ = √2 sin(90°-θ)
sinθ + cosθ = √2 cosθ
sinθ = √2 cosθ - cosθ
sinθ = cosθ(√2 - 1)
sinθ/cosθ = (√2 - 1)
tanθ = (√2 - 1)
cotθ = 1/(√2 - 1) x (√2 + 1)/(√2 + 1)
cotθ = (√2 + 1)

Solution/Answer :
7 sin^2θ + 3cos^2θ = 4
⇒ 7 sin^2θ + 3cos^2θ = 4(sin^2θ + cos^2θ)
⇒ 7 sin^2θ + 3 cos^2θ = 4 sin^2θ + 4 cos^2θ
⇒ 7 sin^2θ − 4sin^2θ = 4 cos^2θ − 3 cos^2θ
⇒ 3 sin^2θ = cos^2θ
⇒ tan^2θ = 1/3
⇒ tanθ = ±1/√3

Solution/Answer :
Given,
secθ + tanθ = m
secθ − tanθ = n
Now,
mn = (secθ + tanθ)(secθ − tanθ)
= sec^2θ−tan^2θ = 1
Thus, mn = 1

Solution/Answer :
x = a secθ + b tanθ
y = a tanθ + b secθ
x^2 − y^2 = a^2 sec^2θ + b^2 tan^2θ + 2ab secθ tanθ − a^2 tan^2θ − b^2 sec^2θ − 2ab secθ tanθ
= a^2 (sec^2θ − tan^2θ) − b^2 (sec^2θ − tan^2θ)
= a^2−b^2 ( As sec^2θ − tan^2θ = 1)
∴ x^2 − y^2 = a^2 − b^2

Solution/Answer :
Given,
x = h + a cosθ
y = k + a sinθ
Now,
x − h = a cosθ
y − k = a sinθ
On squaring and adding we get,
(x − h)^2 + (y − k)^2 = a^2 cos^2θ + a^2 sin^2θ
= a^2 (sin^2θ + cos^2θ)
= a^2(1)
Since [sin^2θ + cos^2θ = 1]
Hence proved.