Introduction to Trigonometry

NCERT Exercise 8.3 Solution part2

Question (3) If tan 2 A = cot (A – 18^o), where 2A is an acute angle, find the value of A.

Solution:

Here, given, tan 2 A = cot (A – 18^o)

We know that, tan A = cot (90^o – A)

∴ tan 2A = cot (90^o – 2A)

Thus, as per question,

90^o – 2A = A – 18^o

⇒ 90^o – 2A + 18^o = A

⇒ 108^o – 2A = A

⇒ A + 2A = 108^o

⇒ 3A = 108^o

∴ A = ^108o/₃

⇒ A = 36^o Answer

Question (4) If tan A = cot B, prove that A + B = 90^o

Solution:

Given, tan A = cot B

We know that, tan A = cot (90^o –A)

Thus, here B = 90^o – A

After transposing A to the left side in the above expression, we get

⇒ B + A = 90^o

⇒ A + B = 90^o Proved

Question (5) If sec 4A = cosec (A – 20^o), where 4A is an acute angle, find the value of A.

Solution:

Given,

sec 4A = cosec (A – 20^o) - - - - - (i)

We know that, sec A = cosec (90^o – A)

∴ sec 4A = cosec (90^o – 4A) - - - - - (ii)

Now, from equation (i) and (ii), we get

90^o – 4A = A – 20^o

After transposing 20^o to LHS and 4A to RHS

⇒ 90^o + 20^o = A + 4A

⇒ 110^o = 5A

⇒ 5A = 110^o

∴ A = ^110o/₅

⇒ A = 22^o Answer

Question (6) If A and B are interior angles of a triangle ABC, then show that sin (^B+C/₂) = cos ^A/₂.

Solution:

Given, A, B, C are interior angles of triangle ABC

Thus, A + B + C = 180^o

⇒ B + C = 180^o – A

After dividing both side by 2, we get

⇒ ^{B + C}/₂ = ^{180o – A}/₂

⇒ ^{B + C}/₂ = ^180o/₂ – ^A/₂

⇒ ^{B + C}/₂ = 90^o – ^A/₂ - - - - (i)

Now, LHS of the question,

= sin (^{B + C}/₂)

= sin (90^o – ^A/₂)

[∵ from equation (i) ^{B + C}/₂ = 90^o – ^A/₂]

= cos ^A/₂

[∵ sin (90^o – A) = cosA]

Thus, sin (^{B + C}/₂) = cos ^A/₂ Proved

Question (7) Express sin 67^o + cos 75^o in terms of trigonometric ratios of angles between 0^o and 45^o.

Solution:

Given,

sin 67^o + cos 75^o

= sin (90^o – 23^o) + cos (90^o – 15^o)

[∵ sin (90^o – A) = cos A and cos (90^o – A) = sin A]

= cos 23^o + sin 15^o Answer

MCQs Test

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