**1. **Evaluate

(i) ^{sin18}^{˚}/_{cos72˚}

(ii) ^{tan26}^{˚}/_{cot64˚}

(iii) cos48˚ – sin42˚

(iv) cosec31˚ – sec59˚

Solution:

(i) sin18˚/cos72˚

= [sin(90˚-72˚)]/cos72˚

= cos72˚)]/cos72˚

=1

(ii) tan26˚/cos64˚

= tan(90˚-64˚)/cos64˚

= cot64˚/cot64˚

= 1

(III) cos 48° − sin 42°

= cos (90°− 42°) − sin 42°

= sin 42° − sin 42°

= 0

(iv) cosec 31° − sec 59°

= cosec (90° − 59°) − sec 59°

= sec 59° − sec 59°

= 0

2. Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II) cos 38° cos 52° − sin 38° sin 52° = 0

(I) tan 48° tan 23° tan 42° tan 67°

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

= 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°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°

= sin 52° sin 38° − sin 38° sin 52°

= 0

**3. If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A. **

Solution:

Given that, tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°) 90° − 2A = A− 18°

108° = 3A

A = 36°

**4. If tan A = cot B, prove that A + B = 90° **

Solution:

Given that, tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°

**5. If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.**

Solution:

Given that, sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

**6. If A, Band C are interior angles of a triangle ABC then show that **

sin[(B+C)/2] = cos A/2

Solution:

We know that for a triangle ABC, ∠ A + ∠B + ∠C = 180°

∠B + ∠C= 180° − ∠A

[∠B + ∠C]/2= [180° − ∠A]/2

sin [∠B + ∠C]/2= sin[180° − ∠A]/2

sin[(B+C)/2] = cos A/2

**7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°. **

Solution:

sin 67° + cos 75°

= sin (90° − 23°) + cos (90° − 15°)

= cos 23° + sin 15°

Yo, Breath Math, My old pappy was scheduled for Engineering (which he couldn’t even spell at the time), but when he flunked Trig, they thought maybe bean counting would suit him better. So, thanks for your efforts, but this is all Greek to the old fool.

hahahaha!

Thank you

Ok