Whakaoti i te sec(90-θ)

$$\sec ( 90 - \theta )$$

Kimi Pārōnaki e ai ki θ

$$-\frac{\tan(90-\theta )}{\cos(90-\theta )}$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Mēnā ko F te hanganga o ngā pānga e rua e taea ana te pārōnaki f\left(u\right) me u=g\left(x\right), arā, mēnā ko F\left(x\right)=f\left(g\left(x\right)\right), ko te pārōnaki o F te pārōnaki o f e ai ki u whakareatia te pārōnaki o g e ai ki x, arā, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$\sec(-\theta ^{1}+90)\tan(-\theta ^{1}+90)\frac{\mathrm{d}}{\mathrm{d}\theta }(-\theta ^{1}+90)$$

Ko te pārōnaki o tētahi pūrau ko te tapeke o ngā pārōnaki o ōna kīanga tau. Ko te pārōnaki o tētahi kīanga tau pūmau ko 0. Ko te pārōnaki o te ax^{n} ko te nax^{n-1}.

$$\sec(-\theta ^{1}+90)\tan(-\theta ^{1}+90)\left(-1\right)\theta ^{1-1}$$

Whakarūnātia.

$$-\sec(-\theta ^{1}+90)\tan(-\theta ^{1}+90)$$

Mō tētahi kupu t, t^{1}=t.

$$-\sec(-\theta +90)\tan(-\theta +90)$$

Aromātai

$$\frac{1}{\cos(90-\theta )}$$