WebSolution Verified by Toppr sec 4A−sec 2A=sec 2A(sec 2A−1) =(1+tan 2A)(1+tan 2A−1) =tan 2A+tan 4A Was this answer helpful? 0 0 Similar questions Solve: (secA+cosA)(secA−cosA)=tan 2A+sin 2A Medium View solution > If sinA+sin 2A+sin 3A=1, then find the value of cos 6A−4cos 4A+8cos 2A. Medium View solution > View more Get … WebProve that:sec8A 1/ sec4A 1 = tan8A/tan2A. Byju's Answer Standard XII Mathematics Algebra of Derivatives Prove that:se... Question Prove that: ( s e c 8 A - 1) ( s e c 4 A - 1) = …
9. Maths Trigonometry Problem Prove that (sec 8A - 1) upon (sec …
Websec4A−1sec8A−1= A 0 B tan2Atan8A C cos2Acos8A D sin2Asin8A Hard Solution Verified by Toppr Correct option is B) Solve any question of Introduction to Trigonometry with:- … Web16 Sep 2024 · tan tan( Let's replace t with a tan (4 Let's take the ratio of (201) and (202) yielding (203), which simplifies to (204). tan) × tan(2a) = exsec8a exsec4a tan(8a) tan(2a) … instep petronas training
If $\\sec(t) = a + 1/(4a)$, prove that $\\sec(t) + \\tan(t) = 2a$ or $1 …
WebSolve: sec4A−1sec8A−1− tan4Atan8A Hard Solution Verified by Toppr Correct option is A) sec4A−1sec8A−1− tan4Atan8A I II Let I= sec4A−1sec8A−1 = cos4A1 −1 cos8A1 −1 ⇒ cos8A1−cos8A× 1−cos4Acos4A ∵cos2θ=2cos 2θ−1=1−2sin 2θ = cos8A1−(1−2sin 24A)× 1−(1−2sin 22A)cos4A = cos8A2sin 24A× 2sin 22Acos4A ∵sin2θ=2sinθcosθ So, ⇒ … WebAnswer (1 of 5): Like this - \frac{1+\tan^2a}{1-\tan^2a} = \sec2A 1+tan^2a is the same as sec^2a. So, \frac{sec^2a}{1-tan^2a} = \sec2a That can be also written as \frac{1}{cos^2} \displaystyle\frac{\frac{1}{cos^2}}{1-\frac{sin^2a}{cos^2a}} = \sec2a The 1 in the denominator, to subtract from... Web9 Dec 2024 · Find an answer to your question Sec 8a_1 ÷ sec 4a _ 1 = tan 8a÷ tan2a. kishankumar8069 kishankumar8069 09.12.2024 Math Secondary School answered • expert verified Sec 8a_1 ÷ sec 4a _ 1 = tan 8a÷ tan2a ... Advertisement abhi178 abhi178 we have to prove that, (sec8a - 1)/(sec4a - 1) = tan8a/tan2a . proof: LHS = (sec8a - 1)/(sec4a - 1) = (1 ... instep pain foot