Advertisements
Advertisements
Question
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
Solution
R.H.S = `(sec^2"A")/("cosec"^2"A")`
= `(1 + tan^2"A")/(1 + cot^2"A")` .....`[(because 1 + tan^2"A" = sec^2"A"),(1 + cot^2"A" = "cosec"^2"A")]`
= `(1 + (sin^2"A")/(cos^2"A"))/(1 + (cos^2"A")/(sin^2"A"))`
= `((cos^2"A" + sin^2"A")/(cos^2"A"))/((sin^2"A" + cos^2"A")/(sin^2"A"))`
= `(1/(cos^2"A"))/(1/(sin^2"A"))` .......[∵ sin2A + cos2A = 1]
= `(sin^2"A")/(cos^2"A")`
= tan2A
= tan A . tan A
= `"tan A"/"cot A"`
= L.H.S
∴ `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
APPEARS IN
RELATED QUESTIONS
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
if `cot theta = sqrt3` find the value of `(cosec^2 theta + cot^2 theta)/(cosec^2 theta - sec^2 theta)`
if `sqrt3 tan theta = 3 sin theta` find the value of `sin^2 theta - cos^2 theta`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Use tables to find sine of 47° 32'
Use tables to find sine of 10° 20' + 20° 45'
Use tables to find cosine of 2° 4’
Use tables to find the acute angle θ, if the value of cos θ is 0.6885
Evaluate:
sin 27° sin 63° – cos 63° cos 27°
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If \[\sec\theta = \frac{13}{12}\], find the values of other trigonometric ratios.
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
Evaluate: cos2 25° - sin2 65° - tan2 45°
The value of tan 72° tan 18° is
The value of tan 1° tan 2° tan 3°…. tan 89° is
If cot( 90 – A ) = 1, then ∠A = ?
`tan 47^circ/cot 43^circ` = 1
