Introduction to Trigonometry

Mathematics Class Tenth

10th-Math-home

NCERT Exercise 8.4 Solution part4

(vi) 1 + sin A/1 – sin A = sec A + tan A

Solution:

Given, LHS = 1 + sin A/1 – sin A

By multiplying with 1 + sin A/1 + sin A we get

= 1 + sin A/1 – sin A × 1 + sinA/1 + sin A

= (1 + sin A)2/12 – sin2A

= (1 + sin A)2/cos2 A

= 1 + sin A/cos A

= 1/cos A + sin A / cos A

= sec A + tan A Proved

(vii) sin θ – 2 sin3 θ/2 cos3 θ – cos θ = tan θ

Solution:

Given, LHS = sin θ – 2 sin3 θ/2 cos3 θ – cos θ

By taking sin θ from numerator and cos θ from denominator as common

= sin θ (1 – 2 sin2 θ)/cos θ (2 cos2 θ –1)

= sin θ (1 – 2 (1 – cos2 θ))/cos θ (2cos2θ – 1)

= sin θ (1 – 2 + 2 cos2 θ)/cos θ (2 cos2θ – 1)

= sin θ (– 1 + 2 cos2 θ)/cos θ (2 cos2θ – 1)

= sin θ (2 cos2 θ – 1)/cos θ (2 cos2 θ – 1)

= sin θ/cos θ

= tan θ Proved

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

Solution:

Given, LHS = (sin A + cosec A)2 + (cos A + sec A)2

= sin2 A + cosec2 A + 2sin A cosec A + cos2 A + sec2 A + 2cosA sec A

= sin2 A + cosec2 A + 2 sin A × 1/sin A + cos2 A + sec2 A + 2 cos A × 1/cos A

= sin2 A + cosec2 A + 2 + cos2 A + sec2 A + 2

= (sin2 A + cos2 A) + 4 + cosec2 A + sec2 A

= 1 + 4 + (1 + cot2 A) + (1 + tan2 A)

[∵ (1 + cot2 A = cosec2 A) and (1 + tan2 A = sec2 A)]

= 5 + 1 + cot2 A + 1 + tan2 A

= 5 + 1 + 1 + cot2 A + tan2 A

= 7 + tan2 A + cot2 A Proved

(ix) (cosec A – sin A) (sec A – cos A) = 1/tan A + cot A

[Hint: Simplify LHS and RHS separately]

Solution:

Given, LHS = (cosec A – sin A) (sec A – cos A)

= (1/sin A – sin A) (1/cosA – cos A)

= (1 – sin2 A/sin A) (1 – cos2A/cos A)

= cos2 A/sin A × sin2 A/cos A

= cos A cos A/sin A × sin A sin A/cos A

= cos A sin A

RHS = 1/tan A + cot A

= 1/sinA/cosA + cosA/sinA

= 1/sin2 A + cos2 A/cos A sin A

= 1/1/cos A sin A

[∵ sin2 A + cos2 A = 1]

= 1 × (cos A sin A)/1

= cos A sin A

Thus, LHS = RHS Proved

(x) (1 + tan2 A/1 + cot2 A) = (1 – tan A/1 – cot A)2 = tan2 A

Solution:

Given, LHS (1 + tan2 A/1 + cot2 A)

= 1 + tan2 A/ 1 + 1/tan2 A

= 1 + tan2 A/ tan2 A + 1/tan2A

= 1 + tan2 A × tan2 A/1 + tan2 A

= tan2 A

Now, given, middle term,

= (1 – tan A/1 – cot A)2

= (1-tanA/1 – 1/tanA)2

= (1 – tan A / tan A – 1/tan A )2

= (1 – tanA × tan A/tan A – 1)2

= (1 – tan A)2 tan2 A/(tan A – 1)2

= (1 + tan2 A – 2 tan A)(tan2A)/tan2 A + 1 – 2 tan A

= (1 + tan2 A – 2 tan A) (tan2A)/1 + tan2 A – 2 tan A

= tan2 A

Thus,

(1 + tan2 A/1 + cot2 A) = (1 – tan A/1 – cot A)2 = tan2 A Proved

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