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Question
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
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Solution
LHS = sin 2A
Putting A = 30° in LHS and RHS., we get
LHS = sin 2 x 30° = sin 60° = `sqrt3/2`
RHS = `(2 xx tan 30°)/(1 + tan^2 30°) = (2 xx 1/sqrt3)/( 1 + (1/sqrt3)^2)`
= `(2/sqrt3)/(1 + 1/3). (2/sqrt3)/(4/3)`
= `(2 xx 3)/(sqrt3 xx 4) = sqrt3/4`
Hence,
LHS = RHS
Hence proved.
RELATED QUESTIONS
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Write the value of tan10° tan 20° tan 70° tan 80° .
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
