Advertisements
Advertisements
प्रश्न
Prove that:
sec (70° – θ) = cosec (20° + θ)
योग
Advertisements
उत्तर
sec (70° – θ) = sec [90° – (20° + θ)] = cosec (20° + θ)
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Find the value of x, if sin 2x = 2 sin 45° cos 45°
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Use tables to find sine of 34° 42'
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
Evaluate:
`2(tan35^@/cot55^@)^2 + (cot55^@/tan35^@)^2 - 3(sec40^@/(cosec50^@))`
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
If tan θ = cot 37°, then the value of θ is
If x and y are complementary angles, then ______.
