Advertisements
Advertisements
Question
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
Solution 1
`(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
= `[cot(90°- 49°)]^2/(tan^2 49°)- 2[sin(90° - 15°)]^2/(cos^2 15°)`
= `(tan^2 49°)/(tan^2 59°) - 2 (cos^2 15°)/(cos^2 15°)`
= 1 - 2
= -1
Solution 2
`(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
= `[cot(90°- 49°)]^2/(tan^2 49°)- 2[sin(90° - 15°)]^2/(cos^2 15°)`
= `(tan^2 49°)/(tan^2 49°) - 2 (cos^2 15°)/(cos^2 15°)`
= 1 - 2
= -1
APPEARS IN
RELATED QUESTIONS
if `tan theta = 12/5` find the value of `(1 + sin theta)/(1 -sin theta)`
Solve.
sin15° cos75° + cos15° sin75°
Use tables to find cosine of 2° 4’
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
Prove that:
`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A)) = 2sec^2(90^@ - A)`
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that cos 3 A = 4 cos3 A – 3 cos A
If \[\sec\theta = \frac{13}{12}\], find the values of other trigonometric ratios.
Solve: 2cos2θ + sin θ - 2 = 0.
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)