Math, asked by mathewjustin3831, 9 months ago

Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin59°+cos56°
(ii) tan65°+cot49°
(iii) sec76°+cosec52°
(iv) cos78°+sec78°
(v) cosec54°+sin72°
(vi) cot85°+cos75°
(vii) sin67°+cos75°

Answers

Answered by nikitasingh79
1

Given :  sin 59° + cos 56°

SOLUTION :   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°  

 

(v) SOLUTION :  

Given : cosec 54° + sin 72°

cosec 54° + sin 72° = cosec (90° - 36°) + sin (90° - 18°)

= sec 36° + Cos 18°

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

Hence, cosec 54° + sin 72° = sec 36° + Cos 18°.

 

(vi) SOLUTION :  

Given : cot 85° + cos 75°

cot 85° + cos 75° = cot (90° - 5°) + cos (90° - 15°)

= tan 5° + sin 15°

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

Hence, cot 85° + cos 75° = tan 5° + sin 15°

 

(vii) SOLUTION :  

Given : sin 67° + cos 75°

sin 67° + cos 75° = sin (90° - 23°) + cos (90° - 15°)

= cos 23° + sin 15°  

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

Hence, sin 67° + cos 75° = cos 23° + sin 15°  

HOPE THIS ANSWER WILL HELP YOU……

 

Some more questions :  

Evaluate the following :

(ix)sin 35° sin 55° − cos 35° cos 55°

(x)tan 48° tan 23° tan 42° tan 67°

(xi)sec50° sin 40° + cos40° cosec 50°

https://brainly.in/question/6620931

 

Evaluate the following :

(v)\frac{tan35^{0}}{cot55^{0}}+\frac{cot78^{0}}{tan12^{0}}-1

(vi)\frac{sec70^{0}}{cosec20^{0}}+\frac{sin59^{0}}{cos31^{0}}

(vii)cosec 31° − sec 59°

(viii)(sin 72° + cos 18°) (sin 72° − cos 18°)

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