Advertisements
Advertisements
प्रश्न
Prove that:
tan (55° - A) - cot (35° + A)
उत्तर
tan (55° - A) - cot (35° + A)
= tan [90° - (55° - A)] - cot (35° + A)
= cot (90° - 55° + A) - cot (35° + A)
= cot (35° + A) - cot (35° + A)
= 0
APPEARS IN
संबंधित प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
if `sin theta = 1/sqrt2` find all other trigonometric ratios of angle θ.
Use tables to find sine of 47° 32'
Evaluate:
cos 40° cosec 50° + sin 50° sec 40°
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
If the angle θ = –45° , find the value of tan θ.
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
The value of tan 1° tan 2° tan 3° ...... tan 89° is
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]