Question

solve this one please​

Answer 1

★ Solution ★

L.H.S = $cot\theta-tan\theta$

$=\frac{cos\theta}{sin\theta}-\frac{sin\theta}{cos\theta}\\=\frac{cos^2\theta-sin^2\theta}{sin\theta cos\theta}---eq(1)\\=\frac{cos^2\theta-(1-cos^2\theta)}{sin\theta cos\theta}\\=\frac{cos^2\theta-1+cos^2\theta}{sin\theta cos\theta}\\=\frac{2cos^2\theta-1}{sin\theta cos\theta}-----eq(2)$

Again from (1) we have,

L.H.S

$\frac{(1-sin^2\theta)-sin^2\theta}{sin\theta cos\theta}\\\frac{1-2sin^2\theta}{sin\theta cos\theta}---eq(3)$

From(2) and (3) we get

$cot\theta-tan\theta=\frac{2cos^2\theta-1}{sin\theta cos\theta}\\=\frac{1-2sin^2\theta}{sin\theta cos\theta}$

$\rule{300}2$

★ HENCE ★ ↓

$=\huge\pink{\fbox{\fbox\frac{1-sin^2\theta}{sin\theta cos\theta} }}$

PROVED√√

Answer 2

Qᴜᴇsᴛɪᴏɴ :-

Prove :- (cotA - tanA) = (2cos²A - 1)/(sinAcosA) = (1 - 2sin²A)/(sinAcosA)

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS,

→ (cotA - tanA)

Putting cotA = (cosA/sinA) & tanA = (sinA/cosA) we get,

→ (cosA/sinA) - (sinA/cosA)

taking LCM now,

→ (cos²A - sin²A)/(sinAcosA) -------- Equation (1)..

Now Putting sin²A = (1 - cos²A) in Numerator we get,

→ [ cos²A - (1 - cos²A) ] / (sinAcosA)

→ [ cos²A + cos²A - 1 ] / (sinAcosA)

→ (2cos²A - 1) / (sinAcosA) = MHS (Proved).

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Now, when we put value of cos²A = (1 - sin²A) in Numerator of Equation (1) we get,

→ [ (1 - sin²A) - sin²A ] / (sinAcosA)

→ [ 1 - sin²A - sin²A ] / (sinAcosA)

→ (1 - 2sin²A) / (sinAcosA) = RHS (Proved).

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Shortcut :-

we can use cos2A = (1 - 2sin²A) = (2cos²A - 1) = (cos²A - sin²A) formula also in Equation (1) to get result right away..

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