Introduction to Trigonometry ⇒⇒ Exercise 8.3

Question 1

Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A.

Solution :

(i)We know that,

$$ sin A = { 1 \over cosec A} $$

$$ sin A = { 1 \over \sqrt{cosec^2 A}} $$

Converting sin A in terms of cot A,

After substituting trigonometric identity

cot ² θ+ 1 = cosec ² θ

$$ sin A = { 1 \over \sqrt{cot^2 A + 1}} $$

(ii)We know that,

$$ sec^2 θ = { 1 + tan^2 θ} $$

Converting sec A in terms of cot A

$$ sec^2 A = { 1 + tan^2 A} $$

$$ sec A = \sqrt{{ 1 + tan^2 A} } $$

$$ sec A = \sqrt{{ 1 + {1 \over cot^2 A} } }$$

$$ sec A = \sqrt{{cot^2 A + 1} \over cot^2 A} $$

(iii)

Converting tan A in terms of cot A

$$ tan A = { 1 \over cot A} $$

Question 2

Write all the other trigonometric ratios of $ \angle $ A in terms of sec A .

Solution :

(i) We know that,

$$ sin^2 A = { 1 -cos^2 A} $$

Converting sin A in terms of sec A,

$$ sin A = { \sqrt{1 -cos^2 A}} $$

$$ sin A = { \sqrt{1 -{1 \over sec^2 A}}} $$

$$ sin A = { \sqrt{{sec^2 A -1 }\over sec^2 A}} $$

$$ sin A = { \sqrt{{sec^2 A -1 }} \over sec A} $$

(ii) Converting cos A in terms of sec A,

$$ cos A = { 1 \over sec A} $$

(iii) We know that,

sec ² θ = 1 + tan ²θ

Converting tan A in terms of sec A,

$$ tan^2 A = { sec^2 A - 1 } $$

$$ tan A = { \sqrt{ sec^2 A - 1 } } $$

(iv) Converting cot A in terms of sec A,

$$ cot A = { 1 \over tan A} $$

$$ cot A = { 1 \over { \sqrt{ sec^2 A - 1 } }} $$

(v) Converting cosec A in terms of sec A,

$$ cosec A = { 1 \over sin A} $$

$$ cosec A = { 1 \over { \sqrt{{sec^2 A -1 }} \over sec A} } $$

$$ cosec A = { sec A \over { \sqrt{sec^2 A -1 }} } $$

Question 3 (i)

Choose the correct option. Justify your choice.
(i) $ 9sec^2 A – 9tan^2 A = $

A) 1         B) 9

C) 8         D) 0

Solution :

$$ 9sec^2 A – 9tan^2 A $$

( After taking 9 as common, we get )

$$ ⇒ 9({sec^2 A – tan^2 A }) $$

[ By identity $ ( {sec^2 A – tan^2 A } = 1) $ ]

$$ ⇒ 9(1) $$

$$ ⇒ 9 $$

Therefore, correct answer is = (B) 9

Question 3 (ii)

Choose the correct option. Justify your choice.
(ii) (1 + tan θ + sec θ) (1 + cotθ – cosecθ) =

A) 0         B) 1

C) 2         D) -1

Solution :

$$(1 + tan θ + sec θ) × (1 + cotθ – cosecθ) $$

$$ ⇒{(1 + { sin θ \over cos θ} + { 1 \over cos θ} )} × {(1 + { cos θ \over sin θ} - { 1 \over sin θ} )} $$

$$ ⇒({{cos θ + sin θ + 1 } \over cos θ } )× ({{ sin θ + cos θ - 1 } \over sin θ } ) $$

$$ ⇒({{[cos θ + sin θ] + 1 } \over cos θ } )× ({{ [sin θ + cos θ ]- 1 } \over sin θ } ) $$

$$ ⇒{{[cos θ + sin θ]^2 - 1^2 } \over {cos θ×sin θ} } $$

( [ a² - b ²] = (a + b) (a - b) )

$$ ⇒{{cos^2 θ + sin^2 θ + 2 sin θ×cos θ - 1} \over {cos θ×sin θ} } $$

( By identity $ ( { sin^2 A + cos^2 A } = 1) $ )

$$ ⇒{{1 + 2 sin θ×cos θ - 1} \over {cos θ×sin θ} } $$

$$ ⇒{{ 2 (sin θ×cos θ) } \over {cos θ×sin θ} } $$

$$ ⇒2 $$

Therefore, correct answer is = (C) 2

Question 3 (iii)

Choose the correct option. Justify your choice.
(iii) $ (sec A + tan A)(1 – sin A) = $

A) sec A         B)sin A

C) cosec A         D) cos A

Solution :

$$ (sec A + tan A)×(1 – sin A) $$

$$ ⇒{({ 1 \over cos A}+ { sin A \over cos A} )}×(1 – sin A) $$

$$ ⇒({{ 1 + sin A}\over cos A} )×(1 – sin A) $$

[ a² - b ²] = (a + b) (a - b)

$$ ⇒{(1^2 – sin^2 A)\over cos A} $$

( By identity $ ( { sin^2 A + cos^2 A } = 1) $ )

$$ ⇒{cos^2 A \over cos A} $$

$$ ⇒{ cos A} $$

Therefore, correct answer is = (D)

Question 3 (iv)

Choose the correct option. Justify your choice.
(iv) $ (1+ tan^2 A)\over (1 – cot^2 A) $ =

A) sec² A         B)-1

C) cot ²A         D) tan² A

Solution :

$$ (1+ tan^2 A)\over (1 + cot^2 A) $$

$$⇒ { sec^2 A\over cosec^2 A }$$

( By identity $ ( {1+ tan^2 A } = sec^2 A) $ )

( By identity $ ( {1 + cot^2 A } = cosec^2 A) $ )

$$ ⇒ {{ 1 \over cos^2 A} \over { 1 \over sin^2 A}} $$

$$ ⇒{{ 1 \over cos^2 A}× { sin^2 A \over 1}} $$

$$ ⇒{ tan^2} $$

Therefore, correct answer is = (D)

Question 4 (i)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i)$$({ cosec θ - cot θ })^2 = {{ 1 - cos θ }\over { 1 + cos θ } }$$

Solution :

$$ ({ cosec θ - cot θ })^2 = {{ 1 - cos θ }\over { 1 + cos θ } } $$

Solving L.H.S. of the above equation :

$$({ cosec θ - cot θ })^2 $$

$$⇒{ ({ { 1 \over sinθ} - { cos θ \over sinθ}})^2 }$$

$$⇒{ ({ { 1 - cos θ } \over sinθ})^2 }$$

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

$$⇒{{ ( { 1 - cos θ })( { 1 - cos θ })} \over (1 - cos^2 θ) }$$

By identity $ ( { sin^2 A + cos^2 A } = 1) $

$$⇒{{ ( { 1 - cos θ })( { 1 - cos θ })} \over { ( { 1 + cos θ })( { 1 - cos θ })} }$$

Using : [ a² - b ²] = (a + b) (a - b)

$$⇒{{ ( { 1 - cos θ })} \over ( { 1 + cos θ }) }$$

$$= R.H.S $$Hence proved.

Question 4 (ii)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(ii)$$ {{ cos A \over {1 + Sin A} } + { {1 + Sin A} \over cos A }} = 2 sec A $$

Solution :

$$ {{ cos A \over {1 + sin A} } + { {1 + sin A} \over cos A }} = 2 sec A $$

Solving L.H.S. of the above equation :

$$ {{ cos A \over {1 + sin A} } + { {1 + sin A} \over cos A }} $$

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

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

By identity $ ( { sin^2 A + cos^2 A } = 1) $

$$⇒ {{ 1 + 1 + 2sin A } \over { ( 1+ sin A) cos A }} $$

$$⇒ {{ 2 + 2sin A } \over { ( 1+ sin A) cos A }} $$

By taking 2 common and simplifying

$$⇒ {{ 2 ( 1+ sin A) } \over { ( 1+ sin A) cos A }} $$

$$⇒ { 2 \over cos A } $$

$$⇒ {2 sec A} $$

$$= R.H.S $$Hence proved.

Question 4 (iii)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iii)$$ {{ tan θ \over {1 - cot θ} } + { {cot θ} \over {1 - tan θ} }} = 1 + secθ × cosecθ $$

Solution :

Solving L.H.S. of the above equation :

$$ {{ tan θ \over {1 - cot θ} } + { {cot θ} \over {1 - tan θ} }} $$

$$ ⇒{ { sin θ \over cos θ} \over { 1 - { cos θ \over sin θ}} } + { { cos θ \over sin θ} \over { 1 - { sin θ \over cos θ}} } $$

$$ ⇒{ { sin θ \over cos θ} \over { {sin θ - cos θ} \over sin θ} } + { { cos θ \over sin θ} \over { {cos θ - sin θ }\over cos θ} } $$

$$ ⇒[{ sin θ \over cos θ} × { sin θ \over {sin θ - cos θ}} ] + [{ cos θ \over sin θ} × { cos θ \over {cos θ - sin θ}} ] $$

Taking (-1) as common in the denominator of the second term of the above expression and simplifying

$$ ⇒[{ sin^2 θ \over { cos θ (sin θ - cos θ)} }] - [{ cos ^2 θ \over { sin θ (sin θ - cos θ)} }] $$

Taking $({ 1 \over {(sin θ - cos θ)} })$ as common

$$ ⇒{ 1 \over {(sin θ - cos θ)} }×[{ sin^2 θ \over cos θ } - { cos ^2 θ \over sin θ }] $$

$$ ⇒{ 1 \over {(sin θ - cos θ)} }×[{{ sin^3 θ - cos ^3 θ} \over cos θ× sin θ}] $$

By identity $ a^3 -b^3 = (a-b) (a^2 + b^2 +ab) $

$$⇒ { 1 \over {(sin θ - cos θ)} }× $$

$$[{{ (sin θ-cos θ) (sin^2 θ + cos^2 θ +sin θcos θ)} \over cos θ× sin θ }] $$

$$ ⇒{{ (sin^2 θ + cos^2 θ +sin θ × cos θ)} \over cos θ× sin θ } $$

By identity $ ( { sin^2 A + cos^2 A } = 1) $

$$ ⇒{{ (1 + sin θ × cos θ)} \over cos θ× sin θ } $$

$$ ⇒{{ 1 \over cos θ× sin θ} + {sin θ× cos θ\over cos θ× sin θ }} $$

$$ ⇒{ 1 \over cos θ× sin θ } + 1 $$

$$ ⇒{ sec θ × cosec θ } + 1 $$

$$= R.H.S $$Hence proved.

Question 4 (iv)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iv)$$ {{1 + sec A} \over sec A } = {sin^2 A \over {1- cos A} } $$

Solution :

Solving L.H.S. of the above equation :

$$ {{1 + sec A} \over sec A } $$

$$⇒ {{1 + { 1 \over cos A } } \over { 1 \over cos A } } $$

$$⇒ {{ cos A + 1 } \over cos A } × { cos A \over 1 } $$

$$⇒ { cos A + 1 } $$

Solving R.H.S. of the above equation :

$${sin^2 A \over {1- cos A}} $$

$$⇒ {{1- cos^2 A} \over {1- cos A}} $$

Using : [ a² - b ²] = (a + b) (a - b)

$$⇒ {{({1- cos A})×({1 + cos A})} \over {1- cos A}} $$

By simplifying

$$⇒ ({1 + cos A}) $$

$$⇒ ({ cos A + 1}) $$

$$ L.H.S. = R.H.S. = cos A + 1. $$Hence proved.

Question 4 (v)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(v) $$ {{cos A - sinA + 1} \over {cos A + sinA - 1} } = cosec A + cot A $$ using the identity $cosec^2 A = 1 + cot^2 A. $

Solution :

Solving L.H.S. of the above equation :

$$ {{cos A - sin A + 1} \over {cos A + sin A - 1} } $$

Dividing each term of numerator and denominator by sin A, we have

$$⇒ {{ {cos A \over sin A }- {sin A \over sin A }+ {1\over sin A }} \over {{cos A\over sin A } + {sinA\over sin A } - {1\over sin A }} } $$

$$⇒ {{cot A - 1 + cosec A} \over {cot A + 1 - cosec A} } $$

We know that $ (cosec^2 A = 1 + cot^2 A) $

Hence Substituting $cosec^2 A -cot^2 A = 1 $ in the equation below

$$⇒ {{cot A + cosec A - (cosec^2 A -cot^2 A)} \over {cot A + 1 - cosec A} } $$

Using : [ a² - b ²] = (a + b) (a - b)

$$⇒ {{cot A + cosec A - [(cosec A +cot A)(cosec A -cot A)]} \over {cot A + 1 - cosec A} } $$

$$⇒ {{(cot A + cosec A ) [1 - (cosec A -cot A)]} \over {cot A + 1 - cosec A} } $$

$$⇒ {{(cot A + cosec A ) (1 - cosec A + cot A)} \over {cot A + 1 - cosec A} } $$

$$⇒ {cot A + cosec A } $$

$$ L.H.S = R.H.S $$Hence proved.

Question 4 (vi)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vi)$$ \sqrt {{1 + sin A}\over {1 - sin A} } = sec A + tan A $$

Solution :

Solving L.H.S. of the above equation :

$$ \sqrt {{1 + sin A}\over {1 - sin A} } $$

By rationalising the denominator, we get;

$$ \sqrt {{{1 + sin A}\over {1 - sin A}} × {{1 + sin A}\over {1 + sin A}} } $$

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

Using : [ a² - b ²] = (a + b) (a - b)

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

$$ {(1 + sin A)} \over {cos A} $$

$$ { 1 \over cos A }+ {sin A \over cos A} $$

By simplifying

$$⇒ sec A + tan A $$

$$ L.H.S = R.H.S $$Hence proved.

Question 4 (vii)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vii)$$ {{ sin θ - 2 sin^3 θ} \over {2 cos^3 θ - cos θ } } = tan θ $$

Solution :

Solving L.H.S. of the above equation :

$$ {{ sin θ - 2 sin^3 θ} \over {2 cos^3 θ - cos θ } } $$

taking common and simplifying

$$ ⇒ {{sin θ × ( 1 - 2 sin^2 θ)} \over { cos θ × (2 cos^2 θ - 1 )} } $$

By identity $( sin^2 A + cos^2 A = 1) $

$$ ⇒ {{sin θ × ( 1 - 2( 1- cos^2 θ))} \over { cos θ × (2 cos^2 θ - 1 )} } $$

$$ ⇒ {{sin θ × ( 1 - 2 + 2cos^2 θ)} \over { cos θ × (2 cos^2 θ - 1 )} } $$

$$ ⇒ {{sin θ × ( 2cos^2 θ - 1 )} \over { cos θ × (2 cos^2 θ - 1 )} } $$

$$ ⇒ {sin θ \over cos θ } $$

$$ ⇒ tan θ $$

$$ L.H.S = R.H.S $$Hence proved.

Question 4 (viii)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(viii)$$ {(sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A } $$

Solution :

Solving L.H.S. of the above equation :

$$ {(sin A + cosec A)^2 + (cos A + sec A)^2 } $$

using $ (a + b )^2 = (a^2 +2ab + b^2) $

$$ ⇒ {(sin^2 A + cosec^2 A + 2sin A Cosec A) + (cos^2 A + sec^2 A + 2cos A secA) } $$

$$⇒ {sin^2 A + cosec^2 A + 2sin A × {1 \over sin A } + cos^2 A + sec^2 A + 2cos A ×{1 \over cos A } } $$

$$⇒ {sin^2 A + cos^2 A + cosec^2 A + sec^2 A + 2 + 2 } $$

$$⇒ {1 + cosec^2 A + sec^2 A + 2 + 2 } $$

$$⇒ {5 + cosec^2 A + sec^2 A } $$

[ By identity $ ( {sec^2 A } = 1 + tan^2 A ) $ ]

[ By identity $ ( {cosec^2 A } = 1+ cot^2 A ) $ ]

$$⇒ {5 + 1+ cot^2 A + 1+tan^2 A } $$

$$⇒ {7 +tan^2 A + cot^2 A } $$

$$ L.H.S = R.H.S $$Hence proved.

Question 4 (ix)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(ix)$$ {(cosec A – sin A) (sec A – cos A) = {1 \over {tan A + cot A} } } $$

Solution :

Solving L.H.S. of the above equation :

$$ (cosec A – sin A) (sec A – cos A) $$

$$⇒ ( {1 \over sin A } – sin A) ×( {1 \over cos A } – cos A) $$

$$ ⇒ ( {{1 – sin^2 A }\over sin A } ) ×({ {1– cos^2 A }\over cos A } ) $$

By identity $( sin^2 A + cos^2 A = 1) $

$$ ⇒ ( {{cos^2 A }\over sin A } ) ×({ {sin^2 A }\over cos A } ) $$

$$ ⇒ { cos A × sin A } $$

Solving R.H.S. of the above equation :

$$ {1 \over {tan A + cot A} } $$

$$ ⇒ {1 \over {{sin A\over cos A } +{cos A\over sin A }} } $$

$$ ⇒ {1 × {{sin Acos A } \over {sin^2 A + cos^2 A } } } $$

By identity $( sin^2 A + cos^2 A = 1) $

$$ ⇒ {sin Acos A \over 1 }$$

$$ L.H.S = R.H.S $$Hence proved.

Question 4 (x)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(x)$$ ({{ 1 + tan^2 A } \over {1 + cot^2 A} }) = ({{ 1 - tan A }\over {1 - cot A}})^2 = {tan^2 A } $$

Solution :

Solving L.H.S. of the above equation :

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

$$⇒ {sec^2 A \over cosec^2 A} $$

$$⇒ {1 \over cos^2 A} × sin^2 A $$

$$ ⇒ tan^2 A $$

Solving middle term of the above equation :

$$[{{ 1 - tan A }\over {1 - cot A}}]^2 $$

$$ ⇒ [{{ 1 - tan A }\over {1 - { 1 \over tan A }}}]^2 $$

$$ ⇒ [{{( 1 - tan A ) }\over { ( tan A - 1 ) }}× tan A ]^2 $$

$$ ⇒ [{{( 1 - tan A ) }\over { -(1 - tan A ) }}× tan A ]^2 $$

$$ ⇒ [- tan A ]^2 $$

$$ ⇒ tan^2 A $$

$$ L.H.S = R.H.S $$Hence proved.