Question

sec⁴theta(1-sin⁴ theta)-2 tan⁴theta=1​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\large\boxed{\sf{sec^4\theta\left(1-sin^4\:\theta\right)-2\:tan^4theta=1}}$

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Well, we can solve this in many ways. I am providing you three different ways. You can use any method

Information related to Topic :

♠ This question belongs to trigonometry concept

♠ We use trignometric identies to solve this question

♠ Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

Method 1 :

$\text{Expand the bracket,}\\\\ \sec ^{4}(A)-\sec ^{4}(A) \sin ^{4}(A)-2 \tan ^{2}(A)=1 \\\\Remember,\\ \sec ^{2}(A)-1=\tan ^{2}(A), \tan ^{2}(A)=\sec ^{2}(A)+1 \\\\ \sec ^{4}(A)-\sec ^{4}(A) \sin ^{4}(A)-2 \sec ^{2}-2=1 \\\\ \sin (x)=\dfrac{\tan (x)}{\sqrt{1+\tan ^{2}(x)}}=\dfrac{\tan (x)}{\sec (x)} \\\\\\ \sec ^{4}(A)-\sec ^{4}(A) \dfrac{\tan (A)^{4}}{\sec (A)^{4}}-2=1 \\\\ \sec ^{4}(A)-\tan ^{4}(A)=1$

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Method 2 :

$\text{Consider the L.H.S,}$

$=\sec ^{4} A\left(1-\sin ^{4} A\right)-2 \tan ^{2} A \\\\ =\sec ^{4} A-\sec ^{4} A \sin ^{4} A-2 \tan ^{2} A \\\\ =\sec ^{4} \mathrm{A}-\dfrac{\sin ^{4} \mathrm{A}}{\cos ^{4} \mathrm{A}}-2 \tan ^{2} \mathrm{A} \\ \\=\sec ^{4} A-\tan ^{4} A-2 \tan ^{2} A \\\\=\left[\left(\sec ^{2} \mathrm{A}\right)^{2}-\left(\tan ^{2}\mathrm{A}\right)^{2}\right]-2 \tan ^{2} \mathrm{A}$

$\text{We know that}\\\\ \sec ^{2} A-\tan ^{2} A=1 \\\\Therefore, =\sec ^{2} A+\tan ^{2} A-2 \tan ^{2} A \\\\ =\sec ^{2} A-\tan ^{2} A \\\\ =1$

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Method 3 :

$\sf{Step\: 1: Break \:1-\sin ^{4} A as a^{2}-b^{2} form.\\ 1-\sin ^{4} A=\left(1-\sin ^{2} A\right)\left(1+\sin ^{2} A\right)=\cos ^{2} A\left(1+\sin ^{2} A\right) form.\\\\Step 2: \sec ^{4} A and \cos ^{2} A multiplies to form \sec ^{2} A (L.H.S.) \\=\sec ^{2} A\left(1+\sin ^{2} A\right)-2 \tan ^{2} A \\\\Step 3: Taking \sec ^{2} A common(L.H.S.) \\=\sec ^{2} A 1+\sin ^{2} A-2 \sin ^{2} A\\=\sec ^{2} A 1-\sin ^{2} A=\sec ^{2} A \cos ^{2} A\\\\=1=\mathrm{R} . \mathrm{H.S} (Proved)}$

Hence Proved !!!

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♣ ᴀᴅᴅɪᴛɪᴏɴᴀʟ ɪɴꜰᴏʀᴍᴀᴛɪᴏɴ :

ꜱᴏᴍᴇ ᴛʀɪɢɴᴏᴍᴇᴛʀɪᴄ ɪᴅᴇɴᴛɪᴛɪᴇꜱ :

Sin θ = 1/Csc θ or Csc θ = 1/Sin θ

Cos θ = 1/Sec θ or Sec θ = 1/Cos θ

Tan θ = 1/Cot θ or Cot θ = 1/Tan θ

sin2 a + cos2 a = 1

1+tan2 a  = sec2 a

cosec2 a = 1 + cot2 a

Tan θ = Sin θ/Cos θ

Cot θ = Cos θ/Sin θ

Sin (-θ) = – Sin θ

Cos (-θ) = Cos θ

Tan (-θ) = – Tan θ

Cot (-θ) = – Cot θ

Sec (-θ) = Sec θ

Csc (-θ) = -Csc θ

Sin (90 – θ) = Cos θ

Cos (90 – θ) = Sin θ

Tan (90 – θ) = Cot θ

Cot ( 90 – θ) = Tan θ

Sec (90 – θ) = Csc θ

Csc (90 – θ) = Sec θ

sin(α+β)=sin(α).cos(β)+cos(α).sin(β)

sin(α–β)=sinα.cosβ–cosα.sinβ

cos(α+β)=cosα.cosβ–sinα.sinβ

cos(α–β)=cosα.cosβ+sinα.sinβ