SOLUTION
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. (i) sin A = $P/H$ , Since H = $√(P^2+B^2)$ $.^{.}.$ $P/{√(P^2+B^2)}$ i.e $1/{√(1 +(B/P)^2)}$ Since cot A = $B/P$ $.^{.}.$ sin A = $1/{√(1+cot^2A)}$ (ii) sec A = $H/B$ = ${√(P^2+B^2)}/B$ = ${P{√(1+(B/P)^2)}}/B$ = ${√(1+(B/P)^2)}/{B/P}$ Since cot A = $B/P$ $.^{.}.$ sec A = ${√(1+cot^2 A)}/{cot A}$ (iii) tan A = $P/B$ i.e tan A = $1/{B/P}$ Since cot A = $B/P$ $.^{.}.$ tan A = $1/{cot A}$2. Write all the other trigonometric ratios of ∠ A in terms of sec A. (i) $sin A$ = $√{sin^2A}$ = $√(1 - cos^2A)$ = $√(1 - {1/{sec^2A}})$ = $√({sec^2A-1}/{sec^2A})$ = $√({sec^2A-1})/{sec A}$ (ii) $cos A$ = $1/{sec A}$ (iii) tanA = $√{tan^2A}$ = $√{sec^2A-1}$ (iv) cosec A =$√{cosec^2A}$ = $√{1+cot^2 A}$ = $√{1+1/{tan^2A}}$ = $√{1+1/{sec^2A - 1}}$ = $√({sec^2A}/{sec^2A - 1})$ = ${sec A}/√{sec^2A - 1}$ (v)cot A = $√{cot^2A}$ = $√{1/{tan^2A}}$ = ${1/√{sec^2A - 1}}$3. Evaluate : (i) ${sin^2 63°+ sin^2 27° }/{cos^2 17° + cos^2 73°}$ (ii) $sin 25° cos 65° + cos 25° sin 65°$ General concept for solving these type of problems. Using these concept i.e sin θ = cos(90-θ) and $sin^2θ+cos^2θ = 1$ , these problem can be solved. (i) ${sin^2 63°+ sin^2 27° }/{cos^2 17° + cos^2 73°}$ = ${sin^2 63°+ cos^2 (90°-27°) }/{cos^2 17° + sin^2 (90°-73°)}$ = ${sin^2 63°+ cos^2 (63°) }/{cos^2 17° + sin^2 (17°)}$ = ${1 }/{1}$ = 1 (ii) $sin 25° cos 65° + cos 25° sin 65°$ = $sin 25° sin(90°-65°) + cos 25° cos(90°-65°)$ = $sin 25° sin(25°) + cos 25° cos(25°)$ = $sin^2 25° + cos^2 25°$ = 14. Choose the correct option. Justify your choice. (i) 9 sec2 A – 9 tan2 A = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) = (A) 0 (B) 1 (C) 2 (D) –1 (iii) (sec A + tan A) (1 – sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A (iv) ${1 + tan^2 A }/{1 + cot^2 A }$ (A) sec2 A (B) –1 (C) cot2 A (D) tan2 A (i) Since $tan^2A = sec^2A-1$ therefore 9 sec2 A – 9 tan2 A = $9sec^2A-9(sec^2A-1)$ = 90 (ii)(1 + tan θ + sec θ) (1 + cot θ – cosec θ) = = (1+${sin θ }/{cos θ }$+$1/{cos θ }$)(1+${cos θ }/{sin θ }$ - $1/{sin θ }$) = $1+{cosθ}/{sinθ}-{1}/{sinθ}+{sinθ}/{cosθ}+{sinθcosθ}/{cosθsinθ}-{sinθ}/{cosθsinθ} +1/{cosθ}+{cosθ}/{cosθsinθ} -1/{cosθsinθ}$ = $2+ {cosθ}/{sinθ} + {sinθ}/{cosθ} - 1/{cosθsinθ}$ = $2+ {cos^2θ+sin^2θ}/{sinθcosθ} - 1/{cosθsinθ}$ = $2+ {1}/{sinθcosθ} - 1/{cosθsinθ}$ = $2$ (iii) (sec A + tan A) (1 – sin A) = $(1/{cosA}+{sinA}/{cosA})(1-sinA)$ = $1/{cosA} -{sinA}/{cosA} +{sinA}/{cosA} -{sin^2A}/{cosA}$ =$1/{cosA}-{sin^2A}/{cosA}$ = $(1-sin^2A)/{cosA}$ =${cos^2A}/{cosA}$ = $cosA$ (iv) ${1 + tan^2 A }/{1 + cot^2 A }$ = ${1 + tan^2 A }/{1 + 1/{tan^2 A} }$ = $({1 + tan^2 A }/{tan^2 A + 1 }) tan^2 A$ = $tan^2 A$5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) ${(cosec θ – cot θ)^2 ={1 − cos θ}/{1 + cos θ}}$ = $(1/{sinθ} - {cosθ}/{sinθ})^2$ = $(1/{sin^2θ} + {cos^2θ}/{sin^2θ} -2(1/{sinθ})({cosθ}/{sinθ}))$ = ${1+cos^2θ-2cosθ}/{sin^2θ}$ = ${(1-cosθ)^2}/{sin^2θ}$ = ${(1-cosθ)^2}/{1-cos^2θ}$ = ${(1-cosθ)^2}/{(1+cosθ)(1-cosθ)}$ =${1 − cos θ}/{1 + cos θ}$ Hence L.H.S = R.H.S(ii) ${cosA}/{(1 + sin A)} + {(1 + sin A) }/{cos A}={2 sec A}$ = ${cos^2A+1+sin^2A+2sinA}/{(1+sinA)(cosA)}$ = ${2+2sinA}/{(1+sinA)(cosA)}$ = ${2(1+sinA)}/{(1+sinA)(cosA)}$ = ${2}/{(cosA)}$ =$2secA$(iii) ${tan θ}/{1− cot θ} + {cot θ}/{1 − tan θ}={1 + sec θcosec θ}$ ${sinθ}/{cosθ}/{1-{cosθ}/{sinθ}} +{cosθ}/{sinθ}/{1-{sinθ}/{cosθ}}$ = ${sin^2θ}/{cosθ(sinθ-cosθ)} + {cos^2θ}/{sinθ(cosθ-sinθ)}$ = ${sin^2θ}/{cosθ(sinθ-cosθ)} - {cos^2θ}/{sinθ(sinθ-cosθ)}$ = ${sin^3θ -cos^3θ}/{cosθsinθ(sinθ-cosθ)} $ Since $(a^3-b^3) =(a-b)(a^2+ab+b^2)$ Therefore, ${(sinθ-cosθ)(sin^2θ+sinθcosθ+cos^2θ)}/{cosθsinθ(sinθ-cosθ)} $ = ${(sin^2θ+sinθcosθ+cos^2θ)}/{cosθsinθ} $ = ${(1+sinθcosθ)}/{cosθsinθ} $ = $1/{cosθsinθ} + {cosθsinθ}/{cosθsinθ}$ = ${1 + sec θ cosec θ}$ Hence L.H.S = R.H.S(iv) ${1 + sec A }/{sec A }={sin^2A}/{1-cos A}$ = ${1+1/{cosA}}/{1/{cosA}}$ = $cosA+1$ = ${(cosA+1)(1-cosA)}/{1-cosA}$ = ${1-cos^2A}/{1-cosA}$ = ${sin^2A}/{1-cosA}$ Hence L.H.S = R.H.S(v) ${cos A – sin A + 1 }/{cos A + sin A – 1}={cosec A + cot A}$ Divide numerator and denominator by sinA. Therefore, ${cos A – sin A + 1 }/{cos A + sin A – 1}$=$({cotA-1+cosecA}/{cotA+1-cosecA})$ = $({cotA+cosecA -1}/{cotA+1-cosecA})$ We have to express 1 in terms of $cot and cosec$, since we know that $1 =cosec^2A-cot^2A$ =$({cotA+cosecA -(cosec^2A-cot^2A)}/{cotA+1-cosecA})$ =$({(cotA+cosecA) -(cosecA-cotA)(cotA+cosecA)}/{cotA+1-cosecA})$ =$({(cotA+cosecA)(1-(cosecA-cotA))}/{cotA+1-cosecA})$ =$({(cotA+cosecA)(1-cosecA+cotA))}/{cotA+1-cosecA})$ =$(cotA+cosecA)$ Hence L.H.S = R.H.S(vi) $√{(1 + sin A)/(1 – sin A)} = {secA +tan A}$ Multiply the numerator and denominator by $(1+sinA)$ we get, $√{(1 + sin A)/(1 – sin A)}$ = $√{(1 + sin A)(1+sinA)}/{(1 – sin A)(1+sinA)}$ = ${√{(1+sinA)^2}/(1-sin^2A)}$ =${√{(1+sinA)^2}/(cos^2A)}$ = ${(1+sinA)}/{cosA}$ = $ 1/{cosA}+{sinA}/{cosA}$ = ${secA +tan A}$(vii) ${ sin θ − 2 sin^3 θ}/{ 2 cos^3 θ – cos θ } =tan θ $ = ${sinθ(1-2sin^2θ)}/{cosθ(2cos^2θ-1)}$ since $(sin^2θ+cos^2θ = 1)$ Therefore =${sinθ(1-2(1-cos^2θ))}/{cosθ(2cos^2θ-1)}$ =${sinθ(1-2+2cos^2θ))}/{cosθ(2cos^2θ-1)}$ =${tanθ(2cos^2θ-1))}/{(2cos^2θ-1)}$ =${tanθ}$(viii) ${(sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 }$ ${(sin A + cosec A)^2 + (cos A + sec A)^2}$ = ${(sinA + 1/{sinA})^2 +(cosA+1/{cosA})^2}$ = $({sin^2A}+ 1/{sin^2A} + 2) + ({cos^2A+1/{cos^2A}+2})$ = $5+ 1/{sin^2A} + 1/{cos^2A}$ = $5+ cosec^2A + sec^2A$ Since $cosec^2A=1+tan^2A$ and $sec^2A = 1+cot^2A$ = $5+1+tan^2A+1+cot^2A$ = $7 + tan^2 A + cot^2$ Hence L.H.S = R.H.S(ix) $(cosec A – sin A)(sec A – cos A)={1}/{tan A + cot A} $ = $(1/{sinA} - sinA)(1/{cosA}-cosA)$ =$({1-sin^2A}/{sinA})({1-cos^2A}/{cosA})$ =$({cos^2A}/{sinA})({sin^2A}/{cosA})$ = ${sinA}{cosA}$ Since $sin^2A+cos^2A=1$ = ${sinAcosA}/{sin^2A+cos^2A}$ = $1/{sin^2A}/{sinAcosA}+{cos^2A}/{sinAcosA}$ = $1/{sinA}/{cosA}+{cosA}/{sinA}$ = ${1}/{tanA+cotA}$ Hence L.H.S = R.H.S(x) $({1 + tan^2 A}/{1 + cot^2A }) $ = $({1 − tan A }/{1 – cot A })^2$ =$tan^2 A$ = $({1+{sin^2A}/{cos^2A}}/{1+{cos^2A}/{sin^2A}})$ = $({(cos^2A+sin^2A)/{cos^2A}}/{(sin^2A+cos^2A)/{sin^2A}})$ = $({sin^2A}/{cos^2A})$ = ${tan^2A}$