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
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°
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Evaluate the following :
(v)
(vi)
(vii)cosec 31° − sec 59°
(viii)(sin 72° + cos 18°) (sin 72° − cos 18°)
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