CBSE 10th Maths Chapter 8 - Introduction to Trigonometry-Exercise 8.3

Solution:

(i) $\dfrac{sin 18°}{cos 72°}$

To simplify this, convert the sin function into cos function

We know that 18° is written as 90° – 18°, which is equal to the cos 72°.

= $\dfrac{sin (90° – 18°) }{cos 72°}$

Substitute the value, to simplify this equation

= $\dfrac{cos 72° }{cos 72°}$ = 1

(ii)$\dfrac{tan 26°}{cot 64°}$

To simplify this, convert the tan function into cot function

We know that, 26° is written as 90° – 26°, which is equal to the cot 64°.

=$\dfrac{ tan (90° – 26°)}{cot 64°}$

Substitute the value, to simplify this equation

= $\dfrac{cot 64°}{cot 64°}$ = 1

(iii) cos 48° – sin 42°

To simplify this, convert the cos function into sin function

We know that, 48° is written as 90° – 42°, which is equal to the sin 42°.

= cos (90° – 42°) – sin 42°

Substitute the value, to simplify this equation

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

To simplify this, convert the cosec function into sec function

We know that, 31° is written as 90° – 59°, which is equal to the sec 59°

= cosec (90° – 59°) – sec 59°

Substitute the value, to simplify this equation

= sec 59° – sec 59° = 0

Solution:

(i) tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

Simplify the given problem by converting some of the cos functions to the sin functions

We know that,

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°

Substitute the values

= sin 52° sin 38° – sin 38° sin 52° = 0

Solution:

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = $\dfrac{108° }{ 3}$

Therefore, the value of A = 36°

Solution:

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

Solution:

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

110° = 5A

A = $\dfrac{110°}{5}$ = 22°

Therefore, the value of A = 22°

Solution:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ $\dfrac{(B+C)}{2}$ = $\dfrac{(180°-A)}{2}$

⇒ $\dfrac{(B+C)}{2}$ = (90°- $\dfrac{A}{2}$)

Now, multiply both sides by sin functions, we get

⇒ $\dfrac{sin (B+C)}{2}$ = sin (90°- $\dfrac{A}{2}$)

Since sin (90°- $\dfrac{A}{2}$) = cos $\dfrac{A}{2}$, the above equation is equal to

sin $\dfrac{(B+C)}{2}$ = cos $\dfrac{A}{2}$

Hence proved.

Solution:

Given:

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

So, sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°