Question

Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i)sin 59° + cos 56°
(ii)tan 65° + cot 49°
(iii)sec 76° + cosec 52°
(iv)cos 78° + sec 78°

Answer 1

(i) SOLUTION :  

Given :  sin 59° + cos 56°

sin 59° + cos 56° = sin (90° - 31°) + cos (90° - 34°)

= cos 31° + sin 34°  

[sin (90 -  θ) = cos  θ , cos (90 - θ) = sin θ]

Hence , sin 59° + cos 56° = cos 31° + sin 34°  

(ii) SOLUTION :  

Given : tan 65° + cot 49°

tan 65° + cot 49° = tan (90° - 25°) + cot (90° - 41°)  

= cot 25° + tan 41°

[tan (90 - θ) = cot θ , cot (90 - θ) = tan θ]

Hence, tan 65° + cot 49° = cot 25° + tan 41°

(iii) SOLUTION :  

Given : sec 76° + cosec 52°

sec 76° + cosec 52° = sec (90° - 14°) + cosec (90° - 38°)

= cosec 14° + sec 38°  

[sec (90 - θ) = cosec θ , cosec (90 - θ) = secθ]

Hence, sec 76° + cosec 52° =  cosec 14° + sec 38°

(iv) SOLUTION :  

Given : cos 78° + sec 78°

cos 78° + sec 78° = cos (90° - 12°) + sec (90° - 12° )

= sin 12° + cosec 12°  

[cos (90 - θ) = sin θ , sec (90 - θ) = cosec θ ]

Hence,  cos 78° + sec 78° = sin 12° + cosec 12°  

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

hey mate

here is your answer

sin a=cos (90-a)

sec a=cosec (90-a)

tan a=cot (90-a)

thus,

1- cos(90-59)+ sin (90-56)

=>cos 31+ sin 34

2- tan 65+ cot 49

=>cot (90-65) +tan (90-49)

=>cot 25+cot 41

3- sec 76+cosec 52

=>cosec (90-76)+sec( 90-52)

=>cosec 14+sec 38

4- cos 78+ sec 78

=>sin (90-78) +cosec (90-78)

=>sin 12+ cosec 12