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प्रश्न
Which is not correct formula?
विकल्प
1 + tan2θ = sec2θ
1 + sec2θ = tan2θ
cosec2θ – cot2θ = 1
sin2θ + cos2θ = 1
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उत्तर
1 + sec2θ = tan2θ
Explanation:
(A) 1 + tan2θ = sec2θ: Correct. This is a fundamental Pythagorean identity.
(B) 1 + sec2θ = tan2θ: Incorrect. Rearranging the correct identity from (A) gives sec2θ – 1 = tan2θ.
(C) cosec2θ – cot2θ = 1: Correct. This is derived from the standard identity 1 + cot2θ = cosec2θ.
(D) sin2θ + cos2θ = 1: Correct. This is the primary Pythagorean trigonometric identity.
APPEARS IN
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following trigonometric identities.
if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
(i)` (1-cos^2 theta )cosec^2theta = 1`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
If x sin3θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ , then show that x2 + y2 = 1.
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
cos 45° = ?
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
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.
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.
